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<time title="创建时间：2019-01-22 09:56:20 / 修改时间：10:46:53" itemprop="dateCreated datePublished" datetime="2019-01-22T09:56:20+08:00">2019-01-22</time>
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<h1 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h1><blockquote>
<p>给定一个字符串，请你找出其中不含有重复字符的 <em>最长子串</em> 的长度<br>示例</p>
<blockquote>
<p>输入: “abcabcbb”<br>输出: 3<br>解释: 因为无重复字符的最长子串是 “abc”，所以其长度为 3。<br>输入: “bbbbb”<br>输出: 1<br>解释: 因为无重复字符的最长子串是 “b”，所以其长度为 1。<br>输入: “pwwkew”<br>输出: 3<br>解释: 因为无重复字符的最长子串是 “wke”，所以其长度为 3。<br> 请注意，你的答案必须是 子串 的长度，”pwke” 是一个子序列，不是子串。 </p>
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<h1 id="我的思路"><a href="#我的思路" class="headerlink" title="我的思路"></a>我的思路</h1><p>拿到题目，感觉像是dp或贪心之类的题目，但毫无头绪。查阅文档后，直接利用string的find函数解题，整体思路比较直接，以每个下标对应字母为子串的头部，直到找到尾部，maxlength记录最大子串的长度。</p>
<h2 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h2><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br></pre></td><td class="code"><pre><span class="line">class Solution {</span><br><span class="line">public:</span><br><span class="line"> int lengthOfLongestSubstring(string s) {</span><br><span class="line"> //maxlength维持最大长度</span><br><span class="line"> //</span><br><span class="line"> if(s.length()==0) return 0;</span><br><span class="line"> int maxlength = 1;</span><br><span class="line"> for(int i=0;i<s.length();i++){</span><br><span class="line"> for(int j=i+1;j<s.length();j++){</span><br><span class="line"> int position = s.find(s[j],i);</span><br><span class="line"> if(position<j){</span><br><span class="line"> break;</span><br><span class="line"> }</span><br><span class="line"> if(maxlength<j-i+1)</span><br><span class="line"> maxlength = j-i+1;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return maxlength;</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure>
<h1 id="别人的代码"><a href="#别人的代码" class="headerlink" title="别人的代码"></a>别人的代码</h1><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line">class Solution {</span><br><span class="line">public:</span><br><span class="line"> int lengthOfLongestSubstring(string s) {</span><br><span class="line"> int index[128];</span><br><span class="line"> for(int k=0;k<128;k++)</span><br><span class="line"> index[k]=0;</span><br><span class="line"> int max=0,n=s.length();</span><br><span class="line"> int i,j;</span><br><span class="line"> for(i=0,j=0;j<n;j++)</span><br><span class="line"> {</span><br><span class="line"> i=(index[s[j]]>i)?index[s[j]]:i;</span><br><span class="line"> max=(max>j-i+1)?max:j-i+1;</span><br><span class="line"> index[s[j]]=j+1;</span><br><span class="line"> }</span><br><span class="line"> return max;</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure>
<h2 id="算法"><a href="#算法" class="headerlink" title="算法"></a>算法</h2><p>j从左向右遍历字符数组，i记录子串左端，index存储扫描过的字符最后一次出现的位置的下一位，那么当一个子串中出现重复字符时（注意这个字符目前不算做扫描过的字符），i的下标会更新为index位，即：对于abcda，当j扫描到第二个a时，i由原来第一个a的位置移动到其下一位，也就是b的位置。<br>与二重循环相比，这种方法减少了不必要的子串扫描，比如对于dabca来说，二重循环会从每一位开始找子串，而一重循环的子串会跳过第一个a。<br>该算法还巧用ASCII码，将每种字符唯一表示。</p>
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<h1 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h1><blockquote>
<p>给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。<br>如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。<br>您可以假设除了数字 0 之外，这两个数都不会以 0 开头。<br>示例：</p>
<blockquote>
<p>输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)<br>输出：7 -> 0 -> 8<br>原因：342 + 465 = 807 </p>
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<h1 id="我的思路"><a href="#我的思路" class="headerlink" title="我的思路"></a>我的思路</h1><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line">list r;//存储相加结果，有一个指向它的head</span><br><span class="line">int jinwei = 0;//保存进位情况</span><br><span class="line">while(l1!=null&&l2!=null){</span><br><span class="line"> r.value = (l1.value+l2.value+jinwei)%10;</span><br><span class="line"> jinwei = (l1.value+l2.value+jinwei)/10;</span><br><span class="line"> l1 = l1.next;</span><br><span class="line"> l2 = l2.next;</span><br><span class="line"> r = r.next;</span><br><span class="line">}</span><br><span class="line">while(l1!=null){</span><br><span class="line"> r.value = (l1.value+jinwei)%10;</span><br><span class="line"> jinwei = (l1.value+jinwei)/10;</span><br><span class="line"> l1 = l1.next;</span><br><span class="line"> r = r.next;</span><br><span class="line">}</span><br><span class="line">while(l2!=null){</span><br><span class="line"> r.value = (l2.value+jinwei)%10;</span><br><span class="line"> jinwei = (l2.value+jinwei)/10;</span><br><span class="line"> l2 = l2.next;</span><br><span class="line"> r = r.next;</span><br><span class="line">}</span><br><span class="line">return head;</span><br></pre></td></tr></table></figure>
<h2 id="遇到的问题"><a href="#遇到的问题" class="headerlink" title="遇到的问题"></a>遇到的问题</h2><p>数据结构方面并不是非常清楚，所以编译起来会有各种各样的语法错误。试过c++和java，都做的不是很好，基础语法有待加强。要学会new!</p>
<h1 id="官方题解"><a href="#官方题解" class="headerlink" title="官方题解"></a>官方题解</h1><p>官方给出的是java的示例代码：<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br></pre></td><td class="code"><pre><span class="line">public ListNode addTwoNumbers(ListNode l1, ListNode l2) {</span><br><span class="line"> ListNode dummyHead = new ListNode(0);</span><br><span class="line"> ListNode p = l1, q = l2, curr = dummyHead;</span><br><span class="line"> int carry = 0;</span><br><span class="line"> while (p != null || q != null) {</span><br><span class="line"> int x = (p != null) ? p.val : 0;</span><br><span class="line"> int y = (q != null) ? q.val : 0;</span><br><span class="line"> int sum = carry + x + y;</span><br><span class="line"> carry = sum / 10;</span><br><span class="line"> curr.next = new ListNode(sum % 10);</span><br><span class="line"> curr = curr.next;</span><br><span class="line"> if (p != null) p = p.next;</span><br><span class="line"> if (q != null) q = q.next;</span><br><span class="line"> }</span><br><span class="line"> if (carry > 0) {</span><br><span class="line"> curr.next = new ListNode(carry);</span><br><span class="line"> }</span><br><span class="line"> return dummyHead.next;</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
<p>阅读官方题解后发现，我的问题在于没有new一个新节点，也没有head,p,q这一类ListNode作为遍历指针。并且在最后没有检查进位值是否大于0。<br>参照实现我的代码如下：<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br></pre></td><td class="code"><pre><span class="line">/**</span><br><span class="line"> * Definition for singly-linked list.</span><br><span class="line"> * public class ListNode {</span><br><span class="line"> * int val;</span><br><span class="line"> * ListNode next;</span><br><span class="line"> * ListNode(int x) { val = x; }</span><br><span class="line"> * }</span><br><span class="line"> */</span><br><span class="line">class Solution {</span><br><span class="line"> public ListNode addTwoNumbers(ListNode l1, ListNode l2) {</span><br><span class="line"> ListNode head = new ListNode(0);</span><br><span class="line"> ListNode curr = head;</span><br><span class="line"> int value = 0;</span><br><span class="line"> int jinwei = 0;</span><br><span class="line"> while(l1!=null&&l2!=null){</span><br><span class="line"> value = (l1.val+l2.val+jinwei)%10;</span><br><span class="line"> jinwei = (l1.val+l2.val+jinwei)/10;</span><br><span class="line"> l1 = l1.next;</span><br><span class="line"> l2 = l2.next;</span><br><span class="line"> curr.next = new ListNode(value);</span><br><span class="line"> curr = curr.next;</span><br><span class="line"> }</span><br><span class="line"> while(l1!=null){</span><br><span class="line"> value = (l1.val+jinwei)%10;</span><br><span class="line"> jinwei = (l1.val+jinwei)/10;</span><br><span class="line"> l1 = l1.next;</span><br><span class="line"> curr.next = new ListNode(value);</span><br><span class="line"> curr = curr.next;</span><br><span class="line"> }</span><br><span class="line"> while(l2!=null){</span><br><span class="line"> value = (l2.val+jinwei)%10;</span><br><span class="line"> jinwei = (l2.val+jinwei)/10;</span><br><span class="line"> l2 = l2.next;</span><br><span class="line"> curr.next = new ListNode(value);</span><br><span class="line"> curr = curr.next;</span><br><span class="line"> }</span><br><span class="line"> if(jinwei>0)</span><br><span class="line"> curr.next = new ListNode(jinwei);</span><br><span class="line"> </span><br><span class="line"> return head.next;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure></p>
<h1 id="链表"><a href="#链表" class="headerlink" title="链表"></a>链表</h1><h2 id="链表的遍历与插入"><a href="#链表的遍历与插入" class="headerlink" title="链表的遍历与插入"></a>链表的遍历与插入</h2><p><em>java</em> </p>
<p>从官方给出的代码中可以看出，想给链表插入一个新node，并且最后返回整个链表。首先需要一个链表头，我们再返回时返回的是它的下一个node。用currentNode遍历链表找到你想要的位置，进行对应操作。<br>java的好处在于不需要折腾指针，*&有段时间不写了我就分不清orz </p>
<p><em>c++</em><br>这里还是想给一下c++的做法，在c++中如何遍历链表以及插入一个新的node？<br>遍历的方式和java类似，要注意最后返回的是指针，插入新的node需要malloc出一个新的Node的空间。<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line">/**</span><br><span class="line"> * Definition for singly-linked list.</span><br><span class="line"> * struct ListNode {</span><br><span class="line"> * int val;</span><br><span class="line"> * ListNode *next;</span><br><span class="line"> * ListNode(int x) : val(x), next(NULL) {}</span><br><span class="line"> * };</span><br><span class="line"> */</span><br><span class="line">class Solution {</span><br><span class="line">public:</span><br><span class="line"> ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {</span><br><span class="line"></span><br><span class="line"> ListNode *p1 = l1;</span><br><span class="line"> ListNode *p2 = l2;</span><br><span class="line"> ListNode head = ListNode(0);</span><br><span class="line"> ListNode *curr = &head;</span><br><span class="line"> int jinwei = 0;</span><br><span class="line"> while(p1!=NULL||p2!=NULL){</span><br><span class="line"> curr->next = (ListNode *)malloc(sizeof(ListNode));</span><br><span class="line"> int x = (p1!=NULL)?(p1->val):0;</span><br><span class="line"> int y = (p2!=NULL)?(p2->val):0;</span><br><span class="line"> int sum = x+y+jinwei;</span><br><span class="line"> jinwei = sum/10;</span><br><span class="line"> curr = curr->next;</span><br><span class="line"> curr->val = sum%10;</span><br><span class="line"> if(p1!=NULL) p1 = p1->next;</span><br><span class="line"> if(p2!=NULL) p2 = p2->next;</span><br><span class="line"> }</span><br><span class="line"> if(jinwei>0){</span><br><span class="line"> curr->next = (ListNode *)malloc(sizeof(ListNode));</span><br><span class="line"> curr = curr->next;</span><br><span class="line"> curr->val = jinwei;</span><br><span class="line"> }</span><br><span class="line"> curr->next = NULL;</span><br><span class="line"> return head.next;</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure></p>
<p>这里要注意在最后给curr->next置为NULL，否则会RE。</p>
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<h1 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h1><p>leetcode-1：两数之和</p>
<blockquote>
<p>给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。<br>你可以假设每种输入只会对应一个答案。但是，你不能重复利用这个数组中同样的元素。</p>
<blockquote>
<p>示例：<br>给定 nums = [2, 7, 11, 15], target = 9<br>因为 nums[0] + nums[1] = 2 + 7 = 9<br>所以返回 [0, 1]</p>
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<h1 id="我的思路"><a href="#我的思路" class="headerlink" title="我的思路"></a>我的思路</h1><h2 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h2><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line">class Solution {</span><br><span class="line">public:</span><br><span class="line"> vector<int> twoSum(vector<int>& nums, int target) {</span><br><span class="line"> //类似算法导论两数之和那道题目</span><br><span class="line"> //首先确定一个元素x</span><br><span class="line"> //再在剩下的元素里找target-x</span><br><span class="line"> //建立一个二维数组存储下标</span><br><span class="line"> vector<int> ans;</span><br><span class="line"> int n = nums.size();</span><br><span class="line"> for(int i=0;i<n-1;i++){</span><br><span class="line"> int x = nums[i];</span><br><span class="line"> int y = target-x;</span><br><span class="line"> for(int j=i+1;j<n;j++){</span><br><span class="line"> if(y==nums[j]){</span><br><span class="line"> ans.push_back(i);</span><br><span class="line"> ans.push_back(j);</span><br><span class="line"> return ans;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return ans;</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure>
<h2 id="时间效率分析"><a href="#时间效率分析" class="headerlink" title="时间效率分析"></a>时间效率分析</h2><p>O(n^2)<br>如果使用二分法，那么是O(n*logn)</p>
<h1 id="别人的代码"><a href="#别人的代码" class="headerlink" title="别人的代码"></a>别人的代码</h1><p>我的代码用时72MS<br>最快的代码用时4MS<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br></pre></td><td class="code"><pre><span class="line">class Solution</span><br><span class="line">{</span><br><span class="line">public:</span><br><span class="line"> // Time : O(n)</span><br><span class="line"> // Space : O(n)</span><br><span class="line"> vector<int> twoSum(vector<int>& nums, int target)</span><br><span class="line"> {</span><br><span class="line"> vector<int> v(2, 0);</span><br><span class="line"> unordered_map<int, int> hash;</span><br><span class="line"> // 从数组最后一个元素开始遍历</span><br><span class="line"> // 进入循环前先 --i</span><br><span class="line"> // hash存储的是value对应的index value ~ index</span><br><span class="line"> for (int i = nums.size(); i--; hash[nums[i]] = i)</span><br><span class="line"> {</span><br><span class="line"> // find查询key是否存在，存在返回其对应迭代器</span><br><span class="line"> if (hash.find(target - nums[i]) == hash.end())</span><br><span class="line"> {</span><br><span class="line"> continue;</span><br><span class="line"> }</span><br><span class="line"> else</span><br><span class="line"> {</span><br><span class="line"> v[0] = i;</span><br><span class="line"> v[1] = hash[target - nums[i]];</span><br><span class="line"> return v;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return v;</span><br><span class="line"> }</span><br><span class="line">};</span><br></pre></td></tr></table></figure></p>
<h2 id="算法"><a href="#算法" class="headerlink" title="算法"></a>算法</h2><p>首先用一个具有两个元素每个元素的初始值为0的vector。<br>再用哈希表来存储已经访问过的元素，元素值作为key，其索引作为value。（这和我们需要比较数字大小这点有关）。<br>对输入的数组只做一次遍历，从后向前，访问过的元素都加入哈希表中，每次检查最新访问的元素是否和哈希表中存在的元素构成满足题意的两数之和。<br>仔细看循环条件，发现这种循环的方式也没有用过，for第二个项使用i–，实际上是if(i–),能在i=0时自动退出循环，减少了代码量。</p>
<h2 id="数据结构"><a href="#数据结构" class="headerlink" title="数据结构"></a>数据结构</h2><h3 id="unordered-map"><a href="#unordered-map" class="headerlink" title="unordered_map"></a>unordered_map</h3><p>C++11中，哈希表对应的容器是unordered_map，适用于需要高效率查询的情况。具体可查阅<a href="http://www.cplusplus.com/reference/unordered_map/unordered_map/" target="_blank" rel="noopener">文档</a>。</p>
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<a href="/2019/01/20/一日一题/" class="post-title-link" itemprop="http://www.kakuketsu.top/index.html">一日一题</a>
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<time title="创建时间：2019-01-20 14:03:33 / 修改时间：14:10:56" itemprop="dateCreated datePublished" datetime="2019-01-20T14:03:33+08:00">2019-01-20</time>
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<span itemprop="about" itemscope="" itemtype="http://schema.org/Thing"><a href="/categories/oj刷题/" itemprop="url" rel="index"><span itemprop="name">oj刷题</span></a></span>
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<p>每日一刷leetcode</p>
<p>之前的计划原本是一日三题</p>
<p>看了leetcode在通过最后给出了执行时间分布表</p>
<p>觉得仅仅一道题目自己也有许多需要学习的部分</p>
<p>所以现在计划是每日至少一题~</p>
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