LeetCode Top 100 Liked Questions 739. Daily Temperatures (Java版； Medium)

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LeetCode Top 100 Liked Questions 739. Daily Temperatures (Java版; Medium)

题目描述

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], 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]

class Solution {
    public int[] dailyTemperatures(int[] T) {
        int n = T.length;
        if(n==0){
            return new int[]{};
        }
        int[] res = new int[n];
        Stack<Integer> s = new Stack<>();
        for(int i=0; i<n; i++){
            while(!s.isEmpty() && T[i]>T[s.peek()]){
                int j = s.pop();
                res[j] = i - j;
            }
            s.push(i);
        }
        //
        return res;
    }
}

第一次做; 单调栈, 遍历阶段+清算阶段

/*
找某个数右边离它最近的比它大的数
很有单调栈的感觉, 不过单调栈更复杂, 可以找到两侧比自己大并且距离自己最近的数
用单调栈的流程解决

单调栈:遍历阶段; 清算阶段
*/
import java.util.Stack;


class Solution {
    public int[] dailyTemperatures(int[] T) {
        //input check
        if(T==null || T.length==0)
            return new int[]{};
        //
        int[] res = new int[T.length];
        //单调栈, 栈中存索引(信息量大!); 栈底到栈顶是递减的
        Stack<Integer> stack = new Stack<>();
        //遍历阶段
        for(int i=0; i<T.length; i++){
            if(stack.isEmpty() || T[i]<=T[stack.peek()])
                //将索引入栈
                stack.push(i);
            else{
                while(!stack.isEmpty() && T[i]>T[stack.peek()]){
                    int index = stack.pop();
                    res[index] = i - index;
                }
                stack.push(i);
            }
        }
        //清算阶段;由于数组默认值是0, 和清算阶段的操作逻辑一样, 所以就不用执行操作了,直接返回结果
        return res;
    }
}