feat: 部分文件增加内容

152.maximum-product-subarray.js

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/*
 * @lc app=leetcode id=152 lang=javascript
 *
 * [152] Maximum Product Subarray
 *
 * https://leetcode.com/problems/maximum-product-subarray/description/
 *
 * algorithms
 * Medium (28.61%)
 * Total Accepted:    202.8K
 * Total Submissions: 700K
 * Testcase Example:  '[2,3,-2,4]'
 *
 * Given an integer array nums, find the contiguous subarray within an array
 * (containing at least one number) which has the largest product.
 *
 * Example 1:
 *
 *
 * Input: [2,3,-2,4]
 * Output: 6
 * Explanation: [2,3] has the largest product 6.
 *
 *
 * Example 2:
 *
 *
 * Input: [-2,0,-1]
 * Output: 0
 * Explanation: The result cannot be 2, because [-2,-1] is not a subarray.
 *
 */
/**
 * @param {number[]} nums
 * @return {number}
 */
var maxProduct = function(nums) {
  // let max = nums[0];
  // let temp = null;
  // for(let i = 0; i < nums.length; i++) {
  //     temp = nums[i];
  //     max = Math.max(temp, max);
  //     for(let j = i + 1; j < nums.length; j++) {
  //         temp *= nums[j];
  //         max = Math.max(temp, max);
  //     }
  // }

  // return max;
  let max = nums[0];
  let min = nums[0];
  let res = nums[0];

  for (let i = 1; i < nums.length; i++) {
    let tmp = min;
    min = Math.min(nums[i], Math.min(max * nums[i], min * nums[i])); // 取最小
    max = Math.max(nums[i], Math.max(max * nums[i], tmp * nums[i])); /// 取最大
    res = Math.max(res, max);
  }
  return res;
};

189.rotate-array.js

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/*
 * @lc app=leetcode id=189 lang=javascript
 *
 * [189] Rotate Array
 *
 * https://leetcode.com/problems/rotate-array/description/
 *
 * algorithms
 * Easy (29.07%)
 * Total Accepted:    287.3K
 * Total Submissions: 966.9K
 * Testcase Example:  '[1,2,3,4,5,6,7]\n3'
 *
 * Given an array, rotate the array to the right by k steps, where k is
 * non-negative.
 *
 * Example 1:
 *
 *
 * Input: [1,2,3,4,5,6,7] and k = 3
 * Output: [5,6,7,1,2,3,4]
 * Explanation:
 * rotate 1 steps to the right: [7,1,2,3,4,5,6]
 * rotate 2 steps to the right: [6,7,1,2,3,4,5]
 * rotate 3 steps to the right: [5,6,7,1,2,3,4]
 *
 *
 * Example 2:
 *
 *
 * Input: [-1,-100,3,99] and k = 2
 * Output: [3,99,-1,-100]
 * Explanation:
 * rotate 1 steps to the right: [99,-1,-100,3]
 * rotate 2 steps to the right: [3,99,-1,-100]
 *
 *
 * Note:
 *
 *
 * Try to come up as many solutions as you can, there are at least 3 different
 * ways to solve this problem.
 * Could you do it in-place with O(1) extra space?
 *
 */
/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function(nums, k) {
  // 就像扩容一样操作
  k = k % nums.length;
  const n = nums.length;

  for (let i = nums.length - 1; i >= 0; i--) {
    nums[i + k] = nums[i];
  }

  for (let i = 0; i < k; i++) {
    nums[i] = nums[n + i];
  }
  nums.length = n;
};

217.contains-duplicate.js

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/*
 * @lc app=leetcode id=217 lang=javascript
 *
 * [217] Contains Duplicate
 *
 * https://leetcode.com/problems/contains-duplicate/description/
 *
 * algorithms
 * Easy (50.92%)
 * Total Accepted:    324K
 * Total Submissions: 628.5K
 * Testcase Example:  '[1,2,3,1]'
 *
 * Given an array of integers, find if the array contains any duplicates.
 * 
 * Your function should return true if any value appears at least twice in the
 * array, and it should return false if every element is distinct.
 * 
 * Example 1:
 * 
 * 
 * Input: [1,2,3,1]
 * Output: true
 * 
 * Example 2:
 * 
 * 
 * Input: [1,2,3,4]
 * Output: false
 * 
 * Example 3:
 * 
 * 
 * Input: [1,1,1,3,3,4,3,2,4,2]
 * Output: true
 * 
 */
/**
 * @param {number[]} nums
 * @return {boolean}
 */
var containsDuplicate = function(nums) {
    // 1. 暴力两层循环两两比较， 时间复杂度O(n^2) 空间复杂度O(1)

    // 2. 先排序，之后比较前后元素是否一致即可，一层循环即可，如果排序使用的比较排序的话时间复杂度O(nlogn) 空间复杂度O(1)

    // 3. 用hashmap ，时间复杂度O(n) 空间复杂度O(n)
    const visited = {};
    for(let i = 0; i < nums.length; i++) {
        if (visited[nums[i]]) return true;
        visited[nums[i]] = true;
    }
    return false;
};

23.merge-k-sorted-lists.js

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/*
 * @lc app=leetcode id=23 lang=javascript
 *
 * [23] Merge k Sorted Lists
 *
 * https://leetcode.com/problems/merge-k-sorted-lists/description/
 *
 * algorithms
 * Hard (33.14%)
 * Total Accepted:    373.7K
 * Total Submissions: 1.1M
 * Testcase Example:  '[[1,4,5],[1,3,4],[2,6]]'
 *
 * Merge k sorted linked lists and return it as one sorted list. Analyze and
 * describe its complexity.
 *
 * Example:
 *
 *
 * Input:
 * [
 * 1->4->5,
 * 1->3->4,
 * 2->6
 * ]
 * Output: 1->1->2->3->4->4->5->6
 *
 *
 */
function mergeTwoLists(l1, l2) {
  const dummyHead = {};
  let current = dummyHead;
  // 1 -> 3 -> 5
  // 2 -> 4 -> 6
  while (l1 !== null && l2 !== null) {
    if (l1.val < l2.val) {
      current.next = l1; // 把小的添加到结果链表
      current = current.next; // 移动结果链表的指针
      l1 = l1.next; // 移动小的那个链表的指针
    } else {
      current.next = l2;
      current = current.next;
      l2 = l2.next;
    }
  }

  if (l1 === null) {
    current.next = l2;
  } else {
    current.next = l1;
  }
  return dummyHead.next;
}
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode[]} lists
 * @return {ListNode}
 */
var mergeKLists = function(lists) {
  // 图参考： https://zhuanlan.zhihu.com/p/61796021
  if (lists.length === 0) return null;
  if (lists.length === 1) return lists[0];
  if (lists.length === 2) {
    return mergeTwoLists(lists[0], lists[1]);
  }

  const mid = lists.length >> 1;
  const l1 = [];
  for (let i = 0; i < mid; i++) {
    l1[i] = lists[i];
  }

  const l2 = [];
  for (let i = mid, j = 0; i < lists.length; i++, j++) {
    l2[j] = lists[i];
  }

  return mergeTwoLists(mergeKLists(l1), mergeKLists(l2));
};

236.lowest-common-ancestor-of-a-binary-tree.js

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/*
 * @lc app=leetcode id=236 lang=javascript
 *
 * [236] Lowest Common Ancestor of a Binary Tree
 *
 * https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
 *
 * algorithms
 * Medium (35.63%)
 * Total Accepted:    267.3K
 * Total Submissions: 729.2K
 * Testcase Example:  '[3,5,1,6,2,0,8,null,null,7,4]\n5\n1'
 *
 * Given a binary tree, find the lowest common ancestor (LCA) of two given
 * nodes in the tree.
 *
 * According to the definition of LCA on Wikipedia: “The lowest common ancestor
 * is defined between two nodes p and q as the lowest node in T that has both p
 * and q as descendants (where we allow a node to be a descendant of itself).”
 *
 * Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]
 *
 *
 *
 * Example 1:
 *
 *
 * Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
 * Output: 3
 * Explanation: The LCA of nodes 5 and 1 is 3.
 *
 *
 * Example 2:
 *
 *
 * Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
 * Output: 5
 * Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant
 * of itself according to the LCA definition.
 *
 *
 *
 *
 * Note:
 *
 *
 * All of the nodes' values will be unique.
 * p and q are different and both values will exist in the binary tree.
 *
 *
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
  if (!root || root === p || root === q) return root;
  const left = lowestCommonAncestor(root.left, p, q);
  const right = lowestCommonAncestor(root.right, p, q);
  if (!left) return right; // 左子树找不到，返回右子树
  if (!right) return left; // 右子树找不到，返回左子树
  return root; // 左右子树分别有一个，则返回root
};

238.product-of-array-except-self.js

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/*
 * @lc app=leetcode id=238 lang=javascript
 *
 * [238] Product of Array Except Self
 *
 * https://leetcode.com/problems/product-of-array-except-self/description/
 *
 * algorithms
 * Medium (53.97%)
 * Total Accepted:    246.5K
 * Total Submissions: 451.4K
 * Testcase Example:  '[1,2,3,4]'
 *
 * Given an array nums of n integers where n > 1,  return an array output such
 * that output[i] is equal to the product of all the elements of nums except
 * nums[i].
 *
 * Example:
 *
 *
 * Input:  [1,2,3,4]
 * Output: [24,12,8,6]
 *
 *
 * Note: Please solve it without division and in O(n).
 *
 * Follow up:
 * Could you solve it with constant space complexity? (The output array does
 * not count as extra space for the purpose of space complexity analysis.)
 *
 */
/**
 * @param {number[]} nums
 * @return {number[]}
 */
var productExceptSelf = function(nums) {
  const ret = [];

  for (let i = 0, temp = 1; i < nums.length; i++) {
    ret[i] = temp;
    temp *= nums[i];
  }
  // 此时ret[i]存放的是前i个元素相乘的结果(不包含第i个)

  // 如果没有上面的循环的话，
  // ret经过下面的循环会变成ret[i]存放的是后i个元素相乘的结果(不包含第i个)

  // 我们的目标是ret[i]存放的所有数字相乘的结果(不包含第i个)

  // 因此我们只需要对于上述的循环产生的ret[i]基础上运算即可
  for (let i = nums.length - 1, temp = 1; i >= 0; i--) {
    ret[i] *= temp;
    temp *= nums[i];
  }
  return ret;
};