【題目】

Given a list of daily temperatures T, return a list such 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 of temperatures T = [73, 74, 75, 71, 69, 72, 76, 73]

Note: The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].

還有幾天會升溫。

https://leetcode.com/problems/daily-temperatures

【思路】

暴力求解，用時太長了……參考答案用stack

【程式碼】

堆疊，倒序比較，當

1、stack不為空，T[i-1]>T[i]，證明不會升溫，pop出棧跳出。進行步驟2

2、判斷stack是否為空，空棧無條件滿足返回0，否則【頂點（升溫點）-迴圈數i 】為天數

3、把i壓入棧。

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

方便理解

class Solution {
    public int[] dailyTemperatures(int[] T) {
        int tmp[]=new int[T.length];
        Arrays.fill(tmp, 1);
        tmp[T.length-1]=0;
        for(int i=0;i<T.length-1;i++){
            int j=i+1;
            while(T[j]<=T[i]){
                    ++j;
                    tmp[i]++;
                if(j>=T.length){
                    break;
                }
            }
            tmp[i]=j<T.length?tmp[i]:0;
        }
        return tmp;
    }
}