[LeetCode] 739. Daily Temperatures

最新推荐文章于 2022-04-19 21:02:51 发布

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最新推荐文章于 2022-04-19 21:02:51 发布

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原文链接：https://segmentfault.com/a/1190000017238993

本文介绍了一种基于栈的数据结构实现的高效算法，用于预测每日气温上升的等待天数。该算法适用于给定一系列每日气温，返回一个列表，对于输入中的每一天，告诉您需要等待多少天才能遇到更温暖的天气。如果未来没有这样的日子，则将结果置为0。

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Problem

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].

Solution

class Solution {
    public int[] dailyTemperatures(int[] T) {
        //use stack to save prev indexes
        int n = T.length;
        int[] res = new int[n];
        Deque<Integer> stack = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            while (!stack.isEmpty() && T[stack.peek()] < T[i]) {
                int preIndex = stack.pop();
                res[preIndex] = i-preIndex;
            }
            stack.push(i);
        }
        return res;
    }
}