Leetcode 739. Daily Temperatures

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原题链接：https://leetcode.com/problems/daily-temperatures/description/

Description：

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].

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Solution：

本题是经典的区间问题，主要就是对每个数求向后的第一个比它大的数的位置，可以引入一个新的的数组，维持一个非递增的序列，当有新的数字加入到该序列时，如果比这个数组末尾的数字大，那么就弹出末位数字，并标记号使之弹出的数字的序号，循环操作，直到如果加入该数字后该序列依然是非递增的序列为止，具体代码如下：

#include <iostream>#include <vector>using namespace std;vector<int> dailyTemperatures(vector<int>& temperatures) {    int len = temperatures.size();    vector<int> temp(len, -1), help;    for (int i = 0; i < len; ++i) {        while (!help.empty() && temperatures[i] > temperatures[help.back()]) {            temp[help.back()] = i;            help.pop_back();        }        help.push_back(i);    }    for (int i = 0; i < len; ++i) {        temp[i] = temp[i] == -1 ? 0 : temp[i] - i;    }    return temp;}int main() {    vector<int> temper = { 73, 74, 75, 71, 69, 72, 76, 73 };    vector<int> res = dailyTemperatures(temper);    for (int i = 0; i < temper.size(); ++i) {        cout << res[i];        if (i < temper.size() - 1) cout << " ";    }    cout << endl;    system("pause");    return 0;}