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Commit 5560742

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feat: daily 2019年06月06日(daily-temperatures) (azl397985856#67)
* feat: longest harmonious subsequence * daily: Longest Harmonious Subsequence, option 1 * pref: make code universal. * pref: get better answer. * alternative solutions. * Update 2019年06月20日.md * Update 2019年06月20日.md * feat: date 2019年06月17日 * fix: add tag. * Update 2019年06月17日.md * Update 2019年06月17日.md * feat: 9.palindrome-number(2019年07月25日) * feat: daily 2019年06月06日(daily-temperatures) * pref: optimize code. * pref: one code for each solution.
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‎daily/2019-06-06.md‎

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# 毎日一题 - 739.Daily Temperatures
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## 信息卡片
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- 时间:2019年06月06日
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- 题目链接:https://leetcode.com/problems/daily-temperatures/
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- tag:`Array` `Stack`
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## 题目描述
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```
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Given a list of daily temperatures T, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.
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For example, given the list of temperatures T = [73, 74, 75, 71, 69, 72, 76, 73], your output should be [1, 1, 4, 2, 1, 1, 0, 0].
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Note: The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].
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```
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## 参考答案
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暴力,双层for循环。`效率很低`
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1. 外层是‘当天’T[i],内层是‘当天’之后T[j];
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2. 多少天之后比‘当天’温度高就是j-i;
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时间复杂度O(n^2), 空间复杂度O(1)
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参考JavaScript代码:
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```js
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/**
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* @param {number[]} T
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* @return {number[]}
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* 双层for循环
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*/
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var dailyTemperatures = function(T) {
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let result = [];
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for(let i = 0; i < T.length; i++) {
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result[i] = 0;
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for(let j = i + 1; j < T.length; j++) {
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if (T[i] < T[j]) {
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result[i] = j - i;
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break;
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}
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}
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}
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return result;
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};
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```
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使用栈,单调递减栈
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1. for循环遍历数组,栈存T的下标i,返回结果数组result;
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2. 拿栈顶元素peek与i比较,T[peek] >= T[i]则将i入栈,T[peek] < T[i]则栈顶值(原数组下标)位置的天数就是result[peek] = i - peek;
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3. 栈顶元素出栈;
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4. 重复2,3两步;
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时间复杂度O(n), 空间复杂度O(n)
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参考JavaScript代码:
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```js
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/**
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* @param {number[]} T
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* @return {number[]}
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* 递减栈;
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*/
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var dailyTemperatures = function(T) {
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let stack = [];
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let result = [];
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for (let i = 0; i < T.length; i++) {
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result[i] = 0;
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while(stack.length > 0 && T[stack[stack.length - 1]] < T[i]) {
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let peek = stack.pop();
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result[peek] = i - peek;
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}
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stack.push(i);
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}
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return result;
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};
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```
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## 优秀解答
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> 暂缺

‎daily/README.md‎

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时间: 2019年06月05日
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#### [739.Daily Temperatures](./2019-06-06.md)
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tag: `Array` `Stack`
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时间: 2019年06月06日
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#### [347.Top K Frequent Elements](./2019-06-08.md)
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tag: `Hash Table` `Heap`

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