[leetcode]739. Daily Temperatures_leetcode 739-优快云博客

博客围绕LeetCode 739题每日温度展开。题目要求根据每日温度列表，计算出等待更暖和天气的天数。博主先用暴力法求解超时，后参考大神解法。还介绍了两种实现方法，一是暴力法，时间复杂度O(n*n)，二是复杂度为O(n)的方法。

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[leetcode]739. Daily Temperatures

Analysis

跟爸妈一起旅游了一个礼拜，然后就回学校准备开学啦~—— [心塞，这么大了还是很讨厌开学]

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].
一开始用暴力解决，然后提交果然超时了，然后参考了一下大神们的解法~具体可以参考一下：https://blog.youkuaiyun.com/kakitgogogo/article/details/78794032

Implement

方法一（暴力,O(n*n)）

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int len = temperatures.size();
        vector<int> res;
        for(int i=0; i<len; i++){
            int cur = temperatures[i];
            int cnt = 0;
            bool flag = false;
            for(int j=i+1; j<len; j++){
                if(temperatures[j] > cur){
                    cnt++;
                    flag = true;
                    break;
                }
                cnt++;
            }
            if(flag)
                res.push_back(cnt);
            else
                res.push_back(0);
        }
        return res;
    }
};

方法二（O(n)）

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int len = temperatures.size();
        vector<int> res(len, 0);
        if(len == 0)
            return res;
        stack<int> index;
        for(int i=0; i<len; i++){
            while(!index.empty() && temperatures[i] > temperatures[index.top()]){
                res[index.top()] = i-index.top();
                index.pop();
            }
            index.push(i);
        }
        return res;
    }
};