leetcode 739. Daily Temperatures 日常温度变化+递减栈Descending Stack

Given a list of daily temperatures, produce a list 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 temperatures = [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].

板梯题意很简单，直接暴力去做，会超时，

实际上这道题应该使用递减栈Descending Stack来做，栈里只有递减元素，思路是这样的，我们遍历数组，如果栈不空，且当前数字大于栈顶元素，那么如果直接入栈的话就不是递减栈了，所以我们取出栈顶元素，那么由于当前数字大于栈顶元素的数字，而且一定是第一个大于栈顶元素的数，那么我们直接求出下标差就是二者的距离了，然后继续看新的栈顶元素，直到当前数字小于等于栈顶元素停止，然后将数字入栈，这样就可以一直保持递减栈，且每个数字和第一个大于它的数的距离也可以算出来了，参见代码如下

代码如下：

#include 
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using namespace std;



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