739.Daily Temperatures
===

###### tags: `Medium`,`Array`,`Stack`

[739. Daily Temperatures](https://leetcode.com/problems/daily-temperatures/)

### 題目描述

Given an array of integers `temperatures` represents the daily temperatures, return an array `answer` such that `answer[i]` is the number of days you have to wait after the i^th^ day to get a warmer temperature. If there is no future day for which this is possible, keep `answer[i] == 0` instead.

### 範例

**Example 1:**

Input: temperatures = [73,74,75,71,69,72,76,73]
Output: [1,1,4,2,1,1,0,0]

**Example 2:**

Input: temperatures = [30,40,50,60]
Output: [1,1,1,0]

**Example 3:**

Input: temperatures = [30,60,90]
Output: [1,1,0]

**Constraints**:

* 1 <= `temperatures.length` <= 10^5^
* 30 <= `temperatures[i]` <= 100

### 解答

#### Javascript

function dailyTemperatures(temperatures) {
  const result = [];
  for (let i = 0; i < temperatures.length; i++) {
    result[i] = 0;
  }
    
  const stack = [];
  for (let i = 0; i < temperatures.length; i++) {
    while (stack.length && temperatures[i] > temperatures[stack[stack.length - 1]]) {
      const current = stack.pop();
      result[current] = i - current;
    }
    stack.push(i); // 只存index就可以了
  }

  return result;
}

>[name=Marsgoat][time= Dec 18, 2022]

#### C#

public class Solution {
    public int[] DailyTemperatures(int[] temperatures) {
        int[] output = new int[temperatures.Length];

        for (int i = temperatures.Length - 1; i >= 0; i--) {
            int j = i + 1;
            while (j < temperatures.Length) {
                if (temperatures[j] > temperatures[i]) {
                    output[i] = j - i;
                    break;
                }

                j += Math.Max(1, output[j]);
            }
        }

        return output;
    }
}

### Reference

[回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)