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

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_ `ith` _day to get a warmer temperature_. If there is no future day for which this is possible, keep `answer[i] == 0` instead.

**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 <= 105`
- `30 <= temperatures[i] <= 100`

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int n = temperatures.size();
        stack<int> st;
        vector<int> res(n);
        for(int i = 0; i < n; i++) {
            // 當遇到溫度是比較高的時候
            while(!st.empty() && temperatures[i] > temperatures[st.top()]) {
                // 拿到前一天的 index
                int previousDay = st.top(); st.pop();
                
                // 計算 index 差，也就是差幾天
                int dayDiff = i - previousDay;
                
                // 將結果存到 res
                res[previousDay] = dayDiff;
            }
            // 將 index 存到 stack
            st.push(i);
        }
        return res;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::