# LeetCode – Daily Temperatures
Date: 2019-07-23


# 题目：

 Daily Temperatures

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.

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]`.

**Note:** The length of `temperatures` will be in the range `[1, 30000]`. Each temperature will be an integer in the range `[30, 100]`.

# 解题：

       使用单调栈如果当前数据比栈顶小则入栈， 如果比栈顶元素大，则出栈并循环处理将所有比当前元素小的栈内元素出栈。

       入栈操作数据为数据索引，出栈时， 计算当前索引与出栈元素（对应数据索引）差值即为元素距离， 设置到对应的vector上并返回。  

# 实现：

```
class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& T) {
        int  n= T.size();
        vector<int> res(n, 0);
        stack<int> ss;
        for (int i=0; i<n; i++){
            while (!ss.empty() && T[i] > T[ss.top()]){
                auto t = ss.top();
                ss.pop();
                res[t] = i - t;
            }
            ss.push(i);
        }
        return res;
    }
};
```