12/18 COMMIT

Author: fupengfei

Leetcode/16. 3Sum Closest.cpp

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+/*3Sum Closest:Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. 
+Return the sum of the three integers. You may assume that each input would have exactly one solution.*/
+class Solution {
+public:
+    int threeSumClosest(vector<int>& nums, int target) {
+        int n = nums.size();
+        if(n < 3) return 0;
+        sort(nums.begin(), nums.end());
+        int res = nums[0] + nums[1] + nums[2];
+        int i, j, k;
+        int val;
+        for (i = 0; i < n - 2; ++i) {
+            j = i + 1;
+            k = n - 1;
+            while (j < k) {
+                val = nums[i] + nums[j] + nums[k];
+                if (abs(val - target) < abs(res - target)) {
+                    res = val;
+                }
+                if (val < target) {
+                    ++j;
+                } else if (val > target) {
+                    --k;
+                } else {
+                    return target;
+                }
+            }
+        }
+        return res;
+    }
+};

Leetcode/17. Letter Combinations of a Phone Number.cpp

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+/*Letter Combinations of a Phone Number:Given a digit string, return all possible letter combinations that the number could represent.*/
+class Solution {
+public:
+	Solution() {
+        letters.push_back("");
+        letters.push_back("");
+        letters.push_back("abc");
+        letters.push_back("def");
+        letters.push_back("ghi");
+        letters.push_back("jkl");
+        letters.push_back("mno");
+        letters.push_back("pqrs");
+        letters.push_back("tuv");
+        letters.push_back("wxyz");
+    }
+    vector<string> letterCombinations(string digits) {
+        vector<string> res;
+        if (digits.size() == 0) {
+            return res;
+        }
+        string s = "";
+        dfs(s, digits, res);
+        return res;
+    }
+private:
+    vector<string> letters;
+    
+    void dfs(string &s, string &digits, vector<string> &res) {
+        int idx = s.size();
+        if (idx == digits.size()) {
+            res.push_back(s);
+            return;
+        }
+        
+        int i;
+        int d = digits[idx] - '0';
+        for (i = 0; i < letters[d].size(); ++i) {
+            s.push_back(letters[d][i]);
+            dfs(s, digits, res);
+            s.pop_back();
+        }
+    }
+};

Leetcode/18. 4Sum.cpp

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+/*4Sum:Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? 
+Find all unique quadruplets in the array which gives the sum of target.*/
+class Solution {
+public:
+    vector<vector<int>> fourSum(vector<int>& nums, int target) {
+        vector<vector<int>> result;
+        int n = nums.size();
+        if(n < 4) return result;
+        sort(nums.begin(), nums.end());
+        for(int i = 0; i < n-3; ++i){
+        	if (i > 0 && nums[i] == nums[i - 1])
+				continue;
+			vector<int> n(nums.begin() + i + 1, nums.end());
+				vector<vector<int>> ret = threeSum(n, target - nums[i]); //基于3sum
+				for (auto v : ret)
+				{
+					v.insert(v.begin(), nums[i]);
+					result.push_back(v);
+				}
+        }
+        return result;
+    }
+    vector<vector<int>> threeSum(vector<int> &num, int target) {
+		vector<vector<int>> result;
+		auto n = num.size();
+		size_t i;
+		for (i = 0; i < n; ++i) {
+			if (i > 0 && num[i] == num[i - 1])
+				continue;
+			int a = num[i];
+			int low = i + 1, high = n - 1;
+			while (low < high) {
+				int b = num[low];
+				int c = num[high];
+				if (a + b + c == target) {
+					vector<int> v = {a, b, c};
+					result.push_back(v);
+					while (low + 1 < n && num[low] == num[low + 1]) low++;
+					while (high - 1 >= 0 && num[high] == num[high - 1]) high--;
+					low++;
+					high--;
+				} else if (a + b + c > target) {
+					while (high - 1 >= 0 && num[high] == num[high - 1]) high--;
+					high--;
+				} else {
+					while (low + 1 < n && num[low] == num[low + 1]) low++;
+					low++;
+				}
+			}
+		}
+		return result;
+	}
+};

Leetcode/19. Remove Nth Node From End of List.cpp

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+/*Remove Nth Node From End of List:Given a linked list, remove the nth node from the end of list and return its head.*/
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+//要求只遍历一趟，需要两个指针p1、p2,p1先走n步，然后p1、p2一起走,直到p1到达链表结尾，此时p2就是需要删除的节点.
+class Solution {
+public:
+    ListNode* removeNthFromEnd(ListNode* head, int n) {
+    	int cnt = n + 1;
+        ListNode* p1 = head;
+        while (p1 != NULL && cnt > 0) {
+            p1 = p1->next;
+            --cnt;
+        }
+        ListNode *tmp;
+        if (cnt > 0) {
+            tmp = head;
+            head = head->next;
+            delete tmp;
+            return head;
+        }
+        ListNode *p2 = head;
+        while (p1 != NULL) {
+            p1 = p1->next;
+            p2 = p2->next;
+        }
+        tmp = p2->next;
+        p2->next = tmp->next;
+        delete tmp;
+        return head;
+    }
+};