[LeetCode] 739. Daily Temperatures

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].

每日温度。题意是给一个数组，表示每天的气温，请返回一个数组，对应位置的输入是你需要再等待多久温度才会升高超过该日的天数。如果之后都不会升高，请在该位置用 0 来代替。例子，

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].

思路依然是单调栈，而且这个题应该是一个单调递增栈。单调栈的题目都会以这个tag展示。创建一个stack，遍历数组，依然是把数组的下标放进stack；若当前stack不为空且栈顶元素背后指向的温度小于试图放进去的index背后的温度，则弹出栈顶index，在结果集里面的对应index可以记录i - index，这就是index和他自己之后一个高温天气的时间差。

时间O(n)

空间O(n)

Java实现

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

JavaScript实现

/**
 * @param {number[]} T
 * @return {number[]}
 */
var dailyTemperatures = function (T) {
    let stack = [];
    let res = new Array(T.length).fill(0);
    for (let i = 0; i < T.length; i++) {
        while (stack.length && T[i] > T[stack[stack.length - 1]]) {
            let index = stack.pop();
            res[index] = i - index;
        }
        stack.push(i);
    }
    return res;
};

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