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剑指 Offer II 038. 每日温度

题目描述

请根据每日 气温 列表 temperatures ，重新生成一个列表，要求其对应位置的输出为：要想观测到更高的气温，至少需要等待的天数。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

 

注意：本题与主站 739 题相同： https://leetcode-cn.com/problems/daily-temperatures/

解法

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)

Python3

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

Java

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int n = temperatures.length;
        int[] res = 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();
                res[j] = i - j;
            }
            stk.push(i);
        }
        return res;
    }
}

C++

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

Go

func dailyTemperatures(temperatures []int) []int {
	res := 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]
			res[j] = i - j
			stk = stk[:len(stk)-1]
		}
		stk = append(stk, i)
	}
	return res
}

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