LeetCode算法题集-739. Daily Temperatures（每日气温）栈的应用

给定一个每日温度的数组，生成一个数组，要求：对于新数组的每个元素，是你需要等待更暖和的天数。如果接下去没有更暖的天了，那就用0替代。

比如，给定数组 temperatures = [73, 74, 75, 71, 69, 72, 76, 73], 你需要输出 [1, 1, 4, 2, 1, 1, 0, 0].原数组第1天是73度，第2天74度是更暖和的一天，所以新数组第1元素就是1.

注: 数组 temperatures 的长度范围： [1, 30000]. 每个温度的范围： [30, 100].

英语原文：739. Daily Temperatures

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].

解答：

class Solution {
public:
    vector dailyTemperatures(vector& temperatures) {
        vector result;
        stack record;
        for (int i=temperatures.size()-1; i>=0; i--)
        {
            while(record.size()>0 && temperatures[record.top()] <= temperatures[i])
            {
                record.pop();
            }
            int wait_day = record.size()==0?0:record.top()-i;
            result.insert(result.begin(), wait_day);
            record.push(i);
        }
        return result;
    }
};