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

清疚 2022-06-03 09:47 148阅读 0赞

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 <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    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;
        }
    };