LeetCode 739. Daily Temperatures

原题链接在这里：https://leetcode.com/problems/daily-temperatures/

题目：

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].

题解：

Stack<Integer> stk里存index, 若当前temperature 更高, 就一直pop出来index, 计算天数差值保存.

最后stk中剩余的index说明这些位置上没有更加暖和的天气, 存0.

Time Complexity: O(n). n = T.length.

Space: O(n).

AC Java:

class Solution {
    public int[] dailyTemperatures(int[] T) {
        if(T == null || T.length == 0){
            return T;
        }
        
        int len = T.length;
        int [] res = new int[len];
        Stack<Integer> stk = new Stack<Integer>();
        for(int i = len-1; i>=0; i--){
            while(!stk.isEmpty() && T[i]>=T[stk.peek()]){
                stk.pop();
            }
            
            res[i] = stk.isEmpty() ? 0 : stk.peek()-i;
            
            stk.push(i);
        }
        
        return res;
    }
}

We could also iterate from left to right. Form a monotonic increasing stack.

When we hit a bigger number, for all the popped indices, we hit a warmer day now. 

Time Complexity: O(n). n = temperatures.size().

Space: O(n).

AC Java:

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        if(temperatures == null || temperatures.length == 0){
            return temperatures;
        }

        int n = temperatures.length;
        Stack<Integer> stk = new Stack<>();
        int [] res = new int[n];
        for(int i = 0; i < n; i++){
            while(!stk.isEmpty() && temperatures[stk.peek()] < temperatures[i]){
                int ind = stk.pop();
                res[ind] = i - ind;
            }

            stk.push(i);
        }
        
        return res;
    }
}

AC C++:

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

            stk.push(i);
        }

        return res;
    }
};

AC Python:

class Solution:
    def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
        n = len(temperatures)
        stk = []
        res = [0] * n
        for i in range(n):
            while stk and temperatures[stk[-1]] < temperatures[i]:
                top = stk.pop()
                res[top] = i - top
            stk.append(i)
        return res

类似Online Stock Span, Next Greater Element I, Minimum Cost Tree From Leaf Values.