LeetCode739 每日温度（Daily temperatures）

题目

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 

方法

单调栈法

遍历数组的数字，将没找到等待时间的数字存入栈中（没找到等待时间意味着截止到遍历的当前数，还没有比它大的数），若遇到比栈中元素大的值（从栈顶开始比较）则记录等待时间并从栈中取出，在栈中未找到等待时间的数字默认等待时间为0

• 时间复杂度：O(n)，n为数组的长度
• 空间复杂度：O(n)，最坏情况下所有数字都要进栈

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