feat: #121 #122 #198 #494 #309 #91

azl397985856 · May 4, 2019 · a86931fe41e959e382d4e02eb92f75b00f22bb03 · a86931f
1 parent edf7916
commit a86931fe41e959e382d4e02eb92f75b00f22bb03
diff --git a/README.md b/README.md
@@ -82,14 +82,17 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 #### 简单难度
-  
- [0020.Valid Parentheses](./problems/20.validParentheses.md) 🖊 
+  
+- [0020.Valid Parentheses](./problems/20.validParentheses.md)
 - [0026.remove-duplicates-from-sorted-array](./problems/26.remove-duplicates-from-sorted-array.md)
 - [0088.merge-sorted-array](./problems/88.merge-sorted-array.md)
+  
+- [0121.best-time-to-buy-and-sell-stock](./problems/121.best-time-to-buy-and-sell-stock.md) 🆕
+  
+- [0122.best-time-to-buy-and-sell-stock-ii](./problems/122.best-time-to-buy-and-sell-stock-ii.md) 🆕
 - [0136.single-number](./problems/136.single-number.md)
 - [0167.two-sum-ii-input-array-is-sorted](./problems/167.two-sum-ii-input-array-is-sorted.md)
 - [0169.majority-element](./problems/169.majority-element.md)
 - [0190.reverse-bits](./problems/190.reverse-bits.md)
 - [0191.number-of-1-bits](./problems/191.number-of-1-bits.md)
+  
+- [0198.house-robber.md](./problems/198.house-robber.md) 🆕
 - [0203.remove-linked-list-elements](./problems/203.remove-linked-list-elements.md)
 - [0206.reverse-linked-list](./problems/206.reverse-linked-list.md)
 - [0219.contains-duplicate-ii](./problems/219.contains-duplicate-ii.md)
@@ -115,7 +118,8 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [0075.sort-colors](./problems/75.sort-colors.md)
 - [0078.subsets](./problems/78.subsets.md) 🆕 
 - [0086.partition-list](./problems/86.partition-list.md)
-  
- [0090.subsets-ii](./problems/90.subsets-ii.md) 🆕 
+  
+- [0090.subsets-ii](./problems/90.subsets-ii.md)
+  
+- [0091.decode-ways](./problems/91.decode-ways.md) 🆕 
 - [0092.reverse-linked-list-ii](./problems/92.reverse-linked-list-ii.md)
 - [0094.binary-tree-inorder-traversal](./problems/94.binary-tree-inorder-traversal.md)
 - [0102.binary-tree-level-order-traversal](./problems/102.binary-tree-level-order-traversal.md)
@@ -131,13 +135,15 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [0236.lowest-common-ancestor-of-a-binary-tree](./problems/236.lowest-common-ancestor-of-a-binary-tree.md)🆕 
 - [0238.product-of-array-except-self](./problems/238.product-of-array-except-self.md) 🆕 
 - [0240.search-a-2-d-matrix-ii](./problems/240.search-a-2-d-matrix-ii.md)
-  
- [0279.perfect-squares](./problems/279.perfect-squares.md)🖊 
+  
+- [0279.perfect-squares](./problems/279.perfect-squares.md)
+  
+- [0309.best-time-to-buy-and-sell-stock-with-cooldown](./problems/309.best-time-to-buy-and-sell-stock-with-cooldown.md) 🆕 
 - [0322.coin-change](./problems/322.coin-change.md)
 - [0334.increasing-triplet-subsequence](./problems/334.increasing-triplet-subsequence.md)
 - [0328.odd-even-linked-list](./problems/328.odd-even-linked-list.md)
 - [0416.partition-equal-subset-sum](./problems/416.partition-equal-subset-sum.md)
 - [0445.add-two-numbers-ii](./problems/445.add-two-numbers-ii.md)
 - [0454.4-sum-ii](./problems/454.4-sum-ii.md) 🆕
+  
+- [0494.target-sum](./problems/494.target-sum.md) 🆕
 - [0518.coin-change-2](./problems/518.coin-change-2.md)
 - [0875.koko-eating-bananas](./problems/875.koko-eating-bananas.md)
 - [0877.stone-game](./problems/877.stone-game.md)

diff --git a/assets/drawio/121.best-time-to-buy-and-sell-stock.drawio b/assets/drawio/121.best-time-to-buy-and-sell-stock.drawio
@@ -0,0 +1 @@
+  
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diff --git a/assets/drawio/122.best-time-to-buy-and-sell-stock-ii.drawio b/assets/drawio/122.best-time-to-buy-and-sell-stock-ii.drawio
@@ -0,0 +1 @@
+  
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diff --git a/assets/drawio/198.house-robber.drawio b/assets/drawio/198.house-robber.drawio
@@ -0,0 +1 @@
+  
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diff --git a/assets/drawio/494.target-sum.drawio b/assets/drawio/494.target-sum.drawio
@@ -0,0 +1 @@
+  
+<mxfile modified="2019-05-04T08:50:46.975Z" host="www.draw.io" agent="Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/73.0.3683.86 Safari/537.36" etag="EDJk86jsYYBlklMIT5Au" version="10.6.6" type="device"><diagram id="4IO8ltCWfnja1ZXnyxj4" name="第 1 页">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</diagram></mxfile>
diff --git a/assets/problems/121.best-time-to-buy-and-sell-stock.png b/assets/problems/121.best-time-to-buy-and-sell-stock.png
diff --git a/assets/problems/122.best-time-to-buy-and-sell-stock-ii.png b/assets/problems/122.best-time-to-buy-and-sell-stock-ii.png
diff --git a/assets/problems/198.house-robber.png b/assets/problems/198.house-robber.png
diff --git a/assets/problems/494.target-sum-2.png b/assets/problems/494.target-sum-2.png
diff --git a/assets/problems/494.target-sum-3.png b/assets/problems/494.target-sum-3.png
diff --git a/assets/problems/494.target-sum.png b/assets/problems/494.target-sum.png
diff --git a/backlog/108.convert-sorted-array-to-binary-search-tree.js b/backlog/108.convert-sorted-array-to-binary-search-tree.js
@@ -0,0 +1,60 @@
+  
+/*
+  
+ * @lc app=leetcode id=108 lang=javascript
+  
+ *
+  
+ * [108] Convert Sorted Array to Binary Search Tree
+  
+ *
+  
+ * https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/
+  
+ *
+  
+ * algorithms
+  
+ * Easy (49.37%)
+  
+ * Total Accepted:    255.2K
+  
+ * Total Submissions: 507.2K
+  
+ * Testcase Example:  '[-10,-3,0,5,9]'
+  
+ *
+  
+ * Given an array where elements are sorted in ascending order, convert it to a
+  
+ * height balanced BST.
+  
+ * 
+  
+ * For this problem, a height-balanced binary tree is defined as a binary tree
+  
+ * in which the depth of the two subtrees of every node never differ by more
+  
+ * than 1.
+  
+ * 
+  
+ * Example:
+  
+ * 
+  
+ * 
+  
+ * Given the sorted array: [-10,-3,0,5,9],
+  
+ * 
+  
+ * One possible answer is: [0,-3,9,-10,null,5], which represents the following
+  
+ * height balanced BST:
+  
+ * 
+  
+ * ⁠     0
+  
+ * ⁠    / \
+  
+ * ⁠  -3   9
+  
+ * ⁠  /   /
+  
+ * ⁠-10  5
+  
+ * 
+  
+ * 
+  
+ */
+  
+/**
+  
+ * Definition for a binary tree node.
+  
+ * function TreeNode(val) {
+  
+ *     this.val = val;
+  
+ *     this.left = this.right = null;
+  
+ * }
+  
+ */
+  
+/**
+  
+ * @param {number[]} nums
+  
+ * @return {TreeNode}
+  
+ */
+  
+var sortedArrayToBST = function(nums) {
+  
+    // 由于数组是排序好的，因此一个思路就是将数组分成两半，一半是左子树，另一半是右子树
+  
+    // 然后运用“树的递归性质”递归完成操作即可。
+  
+    if(nums.length === 0) return null;
+  
+    const mid = nums.length >> 1;
+  
+    const root = new  TreeNode(nums[mid]);
+  
+
+  
+    root.left = sortedArrayToBST(nums.slice(0, mid));
+  
+    root.right = sortedArrayToBST(nums.slice(mid + 1))
+  
+    return root;
+  
+    // 扩展：  这道题启示我们如果是一个非排序的数组，我们可以先进行排序然后再按上述思路进行。
+  
+};
+  
+
diff --git a/backlog/202.happy-number.js b/backlog/202.happy-number.js
@@ -0,0 +1,60 @@
+  
+/*
+  
+ * @lc app=leetcode id=202 lang=javascript
+  
+ *
+  
+ * [202] Happy Number
+  
+ *
+  
+ * https://leetcode.com/problems/happy-number/description/
+  
+ *
+  
+ * algorithms
+  
+ * Easy (44.36%)
+  
+ * Total Accepted:    227.2K
+  
+ * Total Submissions: 505.7K
+  
+ * Testcase Example:  '19'
+  
+ *
+  
+ * Write an algorithm to determine if a number is "happy".
+  
+ * 
+  
+ * A happy number is a number defined by the following process: Starting with
+  
+ * any positive integer, replace the number by the sum of the squares of its
+  
+ * digits, and repeat the process until the number equals 1 (where it will
+  
+ * stay), or it loops endlessly in a cycle which does not include 1. Those
+  
+ * numbers for which this process ends in 1 are happy numbers.
+  
+ * 
+  
+ * Example: 
+  
+ * 
+  
+ * 
+  
+ * Input: 19
+  
+ * Output: true
+  
+ * Explanation: 
+  
+ * 1^2 + 9^2 = 82
+  
+ * 8^2 + 2^2 = 68
+  
+ * 6^2 + 8^2 = 100
+  
+ * 1^2 + 0^2 + 0^2 = 1
+  
+ * 
+  
+ */
+  
+function squareSum(n) {
+  
+    let sum = 0, tmp;
+  
+    while (n) {
+  
+        tmp = n % 10;
+  
+        sum += tmp * tmp;
+  
+        n = Math.floor(n / 10);
+  
+    }
+  
+    return sum;
+  
+}
+  
+
+  
+function isHappyWithMapper(n, visited) {
+  
+    if (n === 1) return true;
+  
+    if (visited[n]) return false;
+  
+    visited[n] = true;
+  
+
+  
+    return isHappyWithMapper(squareSum(n), visited);
+  
+}
+  
+/**
+  
+ * @param {number} n
+  
+ * @return {boolean}
+  
+ */
+  
+var isHappy = function(n) {
+  
+    const visited = {};
+  
+
+  
+    return isHappyWithMapper(n, visited);
+  
+};
+  
+
diff --git a/backlog/204.count-primes.js b/backlog/204.count-primes.js
@@ -0,0 +1,58 @@
+  
+/*
+  
+ * @lc app=leetcode id=204 lang=javascript
+  
+ *
+  
+ * [204] Count Primes
+  
+ *
+  
+ * https://leetcode.com/problems/count-primes/description/
+  
+ *
+  
+ * algorithms
+  
+ * Easy (28.33%)
+  
+ * Total Accepted:    229.8K
+  
+ * Total Submissions: 798.7K
+  
+ * Testcase Example:  '10'
+  
+ *
+  
+ * Count the number of prime numbers less than a non-negative number, n.
+  
+ * 
+  
+ * Example:
+  
+ * 
+  
+ * 
+  
+ * Input: 10
+  
+ * Output: 4
+  
+ * Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
+  
+ * 
+  
+ * 
+  
+ */
+  
+/**
+  
+ * @param {number} n
+  
+ * @return {number}
+  
+ */
+  
+var countPrimes = function(n) {
+  
+    // tag: 数论
+  
+    // if (n <= 2) return 0;
+  
+    // let compositionCount = 0;
+  
+    // for(let i = 3; i < n; i++) {
+  
+    //     for(let j = i - 1; j > 1 ; j--) {
+  
+    //         if (i % j === 0) {
+  
+    //             compositionCount++;
+  
+    //             break; // 找到一个就可以证明它不是质数了
+  
+    //         }
+  
+    //     }
+  
+    // }
+  
+    // return n - compositionCount - 2; // 需要减去1和n这两个数字
+  
+
+  
+
+  
+    // 上面的方法会超时，因此我们需要进行优化
+  
+    // 数学角度来看，如果一个数字可以分解为两个数字相乘（这两个数字不包括0和它本身），那么它就是合数
+  
+    const compositions = []; // compositions[i] 表示i是否是合数
+  
+    let count = 0;
+  
+    for(let i = 2; i < n; i++) {
+  
+        if (!compositions[i]) count++;
+  
+        for(let j = 2; i * j < n; j++) {
+  
+            compositions[i * j] = true;
+  
+        }
+  
+    }
+  
+
+  
+    return count;
+  
+
+  
+};
+  
+
diff --git a/backlog/21.merge-two-sorted-lists.js b/backlog/21.merge-two-sorted-lists.js
@@ -0,0 +1,71 @@
+  
+/*
+  
+ * @lc app=leetcode id=21 lang=javascript
+  
+ *
+  
+ * [21] Merge Two Sorted Lists
+  
+ *
+  
+ * https://leetcode.com/problems/merge-two-sorted-lists/description/
+  
+ *
+  
+ * algorithms
+  
+ * Easy (46.02%)
+  
+ * Total Accepted:    562.7K
+  
+ * Total Submissions: 1.2M
+  
+ * Testcase Example:  '[1,2,4]\n[1,3,4]'
+  
+ *
+  
+ * Merge two sorted linked lists and return it as a new list. The new list
+  
+ * should be made by splicing together the nodes of the first two lists.
+  
+ *
+  
+ * Example:
+  
+ *
+  
+ * Input: 1->2->4, 1->3->4
+  
+ * Output: 1->1->2->3->4->4
+  
+ *
+  
+ *
+  
+ */
+  
+/**
+  
+ * Definition for singly-linked list.
+  
+ * function ListNode(val) {
+  
+ *     this.val = val;
+  
+ *     this.next = null;
+  
+ * }
+  
+ */
+  
+/**
+  
+ * @param {ListNode} l1
+  
+ * @param {ListNode} l2
+  
+ * @return {ListNode}
+  
+ */
+  
+var mergeTwoLists = function(l1, l2) {
+  
+  let current = new ListNode();
+  
+  const dummy = current;
+  
+
+  
+  while (l1 || l2) {
+  
+    if (!l1) {
+  
+      current.next = l2;
+  
+      return dummy.next;
+  
+    } else if (!l2) {
+  
+      current.next = l1;
+  
+      return dummy.next;
+  
+    }
+  
+
+  
+    if (l1.val <= l2.val) {
+  
+      current.next = l1;
+  
+      l1 = l1.next;
+  
+    } else {
+  
+      current.next = l2;
+  
+      l2 = l2.next;
+  
+    }
+  
+
+  
+    current = current.next;
+  
+  }
+  
+
+  
+  return dummy.next;
+  
+
+  
+  //   if (l1 === null) return l2;
+  
+  //   if (l2 === null) return l1;
+  
+  //   if (l1.val < l2.val) {
+  
+  //     l1.next = mergeTwoLists(l1.next, l2);
+  
+  //     return l1;
+  
+  //   } else {
+  
+  //     l2.next = mergeTwoLists(l1, l2.next);
+  
+  //     return l2;
+  
+  //   }
+  
+};
diff --git a/backlog/268.missing-number.js b/backlog/268.missing-number.js
@@ -0,0 +1,50 @@
+  
+/*
+  
+ * @lc app=leetcode id=268 lang=javascript
+  
+ *
+  
+ * [268] Missing Number
+  
+ *
+  
+ * https://leetcode.com/problems/missing-number/description/
+  
+ *
+  
+ * algorithms
+  
+ * Easy (47.60%)
+  
+ * Total Accepted:    267.7K
+  
+ * Total Submissions: 556.2K
+  
+ * Testcase Example:  '[3,0,1]'
+  
+ *
+  
+ * Given an array containing n distinct numbers taken from 0, 1, 2, ..., n,
+  
+ * find the one that is missing from the array.
+  
+ * 
+  
+ * Example 1:
+  
+ * 
+  
+ * 
+  
+ * Input: [3,0,1]
+  
+ * Output: 2
+  
+ * 
+  
+ * 
+  
+ * Example 2:
+  
+ * 
+  
+ * 
+  
+ * Input: [9,6,4,2,3,5,7,0,1]
+  
+ * Output: 8
+  
+ * 
+  
+ * 
+  
+ * Note:
+  
+ * Your algorithm should run in linear runtime complexity. Could you implement
+  
+ * it using only constant extra space complexity?
+  
+ */
+  
+/**
+  
+ * @param {number[]} nums
+  
+ * @return {number}
+  
+ */
+  
+var missingNumber = function(nums) {
+  
+    // 缺失的数字一定是 0 到 n 之间的一个数字
+  
+
+  
+    // 这是一道数论的题目
+  
+    // 这里用到了一条性质： sum([1,n]) = n * (n+1) / 2
+  
+    let sum = 0;
+  
+    for(let num of nums)
+  
+        sum += num;
+  
+
+  
+    return (nums.length * (nums.length + 1) )/ 2 - sum;
+  
+};
+  
+