- https://leetcode-cn.com/problems/reverse-integer/">整数反转 —— https://leetcode-cn.com/problems/reverse-integer/
- https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/">动态规划—股票买入最佳时机问题—https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/
- https://leetcode-cn.com/problems/two-sum/">两数之和—-https://leetcode-cn.com/problems/two-sum/
- 服务器广播 ——https://juejin.im/post/5dc2b45de51d4505277221fc
- 航班预定座位数 ——https://leetcode-cn.com/problems/corporate-flight-bookings/
JS如何判断一组数字是否连续 ——https://segmentfault.com/q/1010000002943744
思路:循环数组,和元素比较=新数组加上此元素,不=则把变量+1继续比较
九宫格按键输入 ——https://blog.csdn.net/shi201619616/article/details/74611161
- 报数游戏 ——https://www.nowcoder.com/questionTerminal/ce46d7af661345f2b1dad0edd491100c
- 字符串分割 ——https://blog.csdn.net/weixin_38169413/article/details/81610165
- 盛最多水的容器 ——https://leetcode-cn.com/problems/container-with-most-water/
- 判读有效字符串 ——https://leetcode-cn.com/problems/valid-parentheses/
整数反转 —— https://leetcode-cn.com/problems/reverse-integer/
/**
* @param {number} x
* @return {number}
*/
var reverse = function(x) {
let reverseX = x.toString().split("").reverse().join("");
reverseX = parseInt(reverseX) * Math.sign(x);
if(Number(reverseX) > Math.pow(2, 31) - 1 || Number(reverseX) < Math.pow(2, 31) * -1) {
return 0;
}
return Number(reverseX);
};
动态规划—股票买入最佳时机问题—https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/
思路:价格比较找到最小价格,记录最大利润,比较目前最大利润和(当前price元素-最小价格),取最大值,一项一项比较到最后就得出最大利润。
/*
* 动态规划
*/
let maxProfit = function(prices) {
let max = 0, minprice = prices[0]
prices.forEach(item => {
minprice = Math.min(item, minprice);
max = Math.max(max, item - minprice)
})
return max
}
两数之和—-https://leetcode-cn.com/problems/two-sum/
思路:构造哈希map,遍历数组,target-item,不存在键值对,则写入哈希map,当前值item,对应下标。 target - item 能在哈希map找到,即满足要求,记录以便函数后面返回
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function(nums, target) {
const targetMap = new Map;
let list = []
nums.forEach((item, index) => {
if(targetMap.has(target - item)) {
list = [index, targetMap.get(target - item)]
} else{
targetMap.set(item, index)
}
});
return list;
};