leetcode hot 04

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收起

解题思路：暴力解法直接遍历一遍，但是最后两个点超时；解决方法用单调栈的方式进行解决，栈存储的是下标，当某一天的温度值比栈顶的温度高的时候，就把栈顶元素弹出，并且在对应answer填入两者下标差值，然后继续对比，直到空栈或者栈顶元素比该天温度高，将该下标存入栈中，然后继续遍历。

暴力解法：

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int[] answer = new int[temperatures.length];
        for(int i=0;i<temperatures.length-1;++i){
            for(int j=i+1;j<temperatures.length;++j){
                if(temperatures[j]>temperatures[i]){
                    answer[i] = j-i;
                    break;
                }
            }
        }
        return answer;
    }
}

栈解法

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        Deque<Integer> stack = new LinkedList<Integer>();
        int[] answer = new int[temperatures.length];
        for(int i=0;i<temperatures.length;i++){
            while(!stack.isEmpty()&&temperatures[stack.peek()]<temperatures[i]){
                answer[stack.peek()] = i-stack.peek();
                stack.pop();
            }
            stack.push(i);
        }
        return answer;
    }
}