977.有序数组的平方 简单题
/*** @param {number[]} nums* @return {number[]}*/
var sortedSquares = function(nums) {const arr = []for(let i = 0; i < nums.length;i++){arr[i] = nums[i]*nums[i]}return arr.sort((a,b)=> a-b)
};
209.长度最小的子数组
考察： 不定滑窗三部曲， 记住要先求得结果， 再更新left
/*** @param {number} target* @param {number[]} nums* @return {number}*/
var minSubArrayLen = function (target, nums) {let ans = Infinitylet left = 0let windowNum = 0for (let left = 0, right = 0; right < nums.length; right++) {// a1 右移做什么windowNum += nums[right]// a2 当满足条件时做什么while (windowNum >= target) {// a3 什么时候更新答案ans = Math.min(ans, right - left + 1)let leftNum = nums[left]windowNum -= leftNumleft += 1}}return ans===Infinity? 0: ans
};
59.螺旋矩阵II
考察点： 边界处理， 左闭右开原则， offset设计， 奇数点
/*** @param {number} n* @return {number[][]}*/
var generateMatrix = function (n) {let startX = 0let startY = 0let count = 1let offset = 1let loop = Math.floor(n / 2)let mid = Math.floor(n / 2)const res = Array(n).fill(0).map(() => Array(n).fill(0))while (loop--) {let row = startXlet col = startY// 左闭右开原则for (; col < n - offset; col++) {res[row][col] = count++}for (; row < n - offset; row++) {res[row][col] = count++}for (; col > startY; col--) {res[row][col] = count++}for (; row > startX; row--) {res[row][col] = count++}startX += 1startY += 1offset += 1}if (n % 2 === 1) {res[mid][mid] = count}return res
};
