daily-temperatures

739. Daily Temperatures

Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead.

Example 1:

Input: temperatures = [73,74,75,71,69,72,76,73] Output: [1,1,4,2,1,1,0,0]

Example 2:

Input: temperatures = [30,40,50,60] Output: [1,1,1,0]

Example 3:

Input: temperatures = [30,60,90] Output: [1,1,0]

Constraints:

• 1 <= temperatures.length <= 105
• 30 <= temperatures[i] <= 100

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int n = temperatures.size();
        stack<int> st;
        vector<int> res(n);
        for(int i = 0; i < n; i++) {
            // 當遇到溫度是比較高的時候
            while(!st.empty() && temperatures[i] > temperatures[st.top()]) {
                // 拿到前一天的 index
                int previousDay = st.top(); st.pop();

                // 計算 index 差，也就是差幾天
                int dayDiff = i - previousDay;

                // 將結果存到 res
                res[previousDay] = dayDiff;
            }
            // 將 index 存到 stack
            st.push(i);
        }
        return res;
    }
};

• T: $O(N)$
• S: $O(N)$