739. Daily Temperatures（python+cpp）

题目：

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].

解释：
用栈做，维护一个降序的栈，栈中存储的是index，index本身不是降序的，但是index对应的T[index]是降序的，当出现比栈中一些index对应的T[index]大的T[i]的时候，把那些index出栈，并保存好位置差，然后把当前的i压入栈，这样就可以实现要求。
python代码：

class Solution:
    def dailyTemperatures(self, T):
        """
        :type T: List[int]
        :rtype: List[int]
        """
        ans=[0]*len(T)
        stack=[]
        for i in range(len(T)):
            while stack and T[stack[-1]]<T[i]:
                cur=stack.pop()
                ans[cur]=i-cur
            stack.append(i)
        return ans

c++代码：

#include <stack>
using namespace std;
class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& T) {
        stack<int>_stack;
        int n=T.size();
        vector<int>result(n,0);
        for(int i=0;i<n;i++)
        {
            while(!_stack.empty() && T[_stack.top()]<T[i])
            {
                int cur=_stack.top();
                _stack.pop();
                result[cur]=i-cur;
            }
            _stack.push(i);
        }
        return result;
    }
};

总结：
while的时候可以多加一些判断。