LeetCode 739. Daily Temperatures

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/Dylan-Java-NYC/p/8192500.html

本文提供了一种使用栈的数据结构解决LeetCode上每日温度问题的有效算法。该算法通过跟踪温度变化并计算直到更暖和天气出现所需的天数来生成答案数组。详细介绍了算法步骤，并附带了完整的Java代码实现。

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

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：

类似Next Greater Element I.

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:

 1 class Solution {
 2     public int[] dailyTemperatures(int[] T) {
 3         if(T == null || T.length == 0){
 4             return T;
 5         }
 6         
 7         int len = T.length;
 8         int [] res = new int[len];
 9         Stack<Integer> stk = new Stack<Integer>();
10         for(int i = len-1; i>=0; i--){
11             while(!stk.isEmpty() && T[i]>=T[stk.peek()]){
12                 stk.pop();
13             }
14             
15             res[i] = stk.isEmpty() ? 0 : stk.peek()-i;
16             
17             stk.push(i);
18         }
19         
20         return res;
21     }
22 }