comments	difficulty	edit_url	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/0700-0799/0739.Daily%20Temperatures/README.md	栈 数组 单调栈

739. 每日温度

English Version

题目描述

给定一个整数数组 temperatures ，表示每天的温度，返回一个数组 answer ，其中 answer[i] 是指对于第 i 天，下一个更高温度出现在几天后。如果气温在这之后都不会升高，请在该位置用 0 来代替。

 

示例 1:

输入: temperatures = [73,74,75,71,69,72,76,73]
输出: [1,1,4,2,1,1,0,0]

示例 2:

输入: temperatures = [30,40,50,60]
输出: [1,1,1,0]

示例 3:

输入: temperatures = [30,60,90]
输出: [1,1,0]

 

提示：

• 1 <= temperatures.length <= 105
• 30 <= temperatures[i] <= 100

解法

方法一：单调栈

单调栈常见模型：找出每个数左/右边离它最近的且比它大/小的数。模板：

stk = []
for i in range(n):
    while stk and check(stk[-1], i):
        stk.pop()
    stk.append(i)

对于本题，我们需要找出每个数右边离它最近的且比它大的数，因此我们可以从右往左遍历数组，且需要维护一个从栈底到栈顶单调递减的栈。

对于当前遍历到的数 temperatures[i]，如果栈顶元素 temperatures[stk[-1]] 小于等于 temperatures[i]，则弹出栈顶元素，直到栈为空或者栈顶元素大于 temperatures[i]。如果此时栈不为空，那么栈顶元素就是 temperatures[i] 右边离它最近的且比它大的数，更新 ans[i] = stk[-1] - i。接着，我们将 $i$ 入栈，继续遍历下一个数。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为 temperatures 数组的长度。

Python3

class Solution:
    def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
        ans = [0] * len(temperatures)
        stk = []
        for i, t in enumerate(temperatures):
            while stk and temperatures[stk[-1]] < t:
                j = stk.pop()
                ans[j] = i - j
            stk.append(i)
        return ans

Java

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int n = temperatures.length;
        int[] ans = new int[n];
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            while (!stk.isEmpty() && temperatures[stk.peek()] < temperatures[i]) {
                int j = stk.pop();
                ans[j] = i - j;
            }
            stk.push(i);
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int n = temperatures.size();
        vector<int> ans(n);
        stack<int> stk;
        for (int i = 0; i < n; ++i) {
            while (!stk.empty() && temperatures[stk.top()] < temperatures[i]) {
                ans[stk.top()] = i - stk.top();
                stk.pop();
            }
            stk.push(i);
        }
        return ans;
    }
};

Go

func dailyTemperatures(temperatures []int) []int {
	ans := make([]int, len(temperatures))
	var stk []int
	for i, t := range temperatures {
		for len(stk) > 0 && temperatures[stk[len(stk)-1]] < t {
			j := stk[len(stk)-1]
			ans[j] = i - j
			stk = stk[:len(stk)-1]
		}
		stk = append(stk, i)
	}
	return ans
}

TypeScript

function dailyTemperatures(temperatures: number[]): number[] {
    const n = temperatures.length;
    const ans = new Array(n).fill(0);
    const stk: number[] = [];
    for (let i = n - 1; i >= 0; --i) {
        while (stk.length && temperatures[stk[stk.length - 1]] <= temperatures[i]) {
            stk.pop();
        }
        if (stk.length) {
            ans[i] = stk[stk.length - 1] - i;
        }
        stk.push(i);
    }
    return ans;
}

Rust

impl Solution {
    pub fn daily_temperatures(temperatures: Vec<i32>) -> Vec<i32> {
        let n = temperatures.len();
        let mut stack = vec![];
        let mut res = vec![0; n];
        for i in 0..n {
            while !stack.is_empty() && temperatures[*stack.last().unwrap()] < temperatures[i] {
                let j = stack.pop().unwrap();
                res[j] = (i - j) as i32;
            }
            stack.push(i);
        }
        res
    }
}

JavaScript

/**
 * @param {number[]} temperatures
 * @return {number[]}
 */
var dailyTemperatures = function (temperatures) {
    const n = temperatures.length;
    const ans = new Array(n).fill(0);
    const stk = [];
    for (let i = n - 1; i >= 0; --i) {
        while (stk.length && temperatures[stk[stk.length - 1]] <= temperatures[i]) {
            stk.pop();
        }
        if (stk.length) {
            ans[i] = stk[stk.length - 1] - i;
        }
        stk.push(i);
    }
    return ans;
};

方法二

Python3

class Solution:
    def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
        n = len(temperatures)
        stk = []
        ans = [0] * n
        for i in range(n - 1, -1, -1):
            while stk and temperatures[stk[-1]] <= temperatures[i]:
                stk.pop()
            if stk:
                ans[i] = stk[-1] - i
            stk.append(i)
        return ans

Java

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int n = temperatures.length;
        Deque<Integer> stk = new ArrayDeque<>();
        int[] ans = new int[n];
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.isEmpty() && temperatures[stk.peek()] <= temperatures[i]) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                ans[i] = stk.peek() - i;
            }
            stk.push(i);
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int n = temperatures.size();
        vector<int> ans(n);
        stack<int> stk;
        for (int i = n - 1; ~i; --i) {
            while (!stk.empty() && temperatures[stk.top()] <= temperatures[i]) stk.pop();
            if (!stk.empty()) ans[i] = stk.top() - i;
            stk.push(i);
        }
        return ans;
    }
};

Go

func dailyTemperatures(temperatures []int) []int {
	n := len(temperatures)
	ans := make([]int, n)
	var stk []int
	for i := n - 1; i >= 0; i-- {
		for len(stk) > 0 && temperatures[stk[len(stk)-1]] <= temperatures[i] {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			ans[i] = stk[len(stk)-1] - i
		}
		stk = append(stk, i)
	}
	return ans
}