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面试题 31. 栈的压入、弹出序列

题目描述

输入两个整数序列，第一个序列表示栈的压入顺序，请判断第二个序列是否为该栈的弹出顺序。假设压入栈的所有数字均不相等。例如，序列 {1,2,3,4,5} 是某栈的压栈序列，序列 {4,5,3,2,1} 是该压栈序列对应的一个弹出序列，但 {4,3,5,1,2} 就不可能是该压栈序列的弹出序列。

 

示例 1：

输入：pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
输出：true
解释：我们可以按以下顺序执行：
push(1), push(2), push(3), push(4), pop() -> 4,
push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1

示例 2：

输入：pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
输出：false
解释：1 不能在 2 之前弹出。

 

提示：

  1. 0 <= pushed.length == popped.length <= 1000
  2. 0 <= pushed[i], popped[i] < 1000
  3. pushed 是 popped 的排列。

注意：本题与主站 946 题相同：https://leetcode.cn/problems/validate-stack-sequences/

解法

借助一个辅助栈实现。

Python3

class Solution:
    def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool:
        s = []
        q = 0
        for num in pushed:
            s.append(num)
            while s and s[-1] == popped[q]:
                s.pop()
                q += 1
        return not s

Java

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        Deque<Integer> s = new ArrayDeque<>();
        int q = 0;
        for (int num : pushed) {
            s.push(num);
            while (!s.isEmpty() && s.peek() == popped[q]) {
                s.pop();
                ++q;
            }
        }
        return s.isEmpty();
    }
}

JavaScript

/**
 * @param {number[]} pushed
 * @param {number[]} popped
 * @return {boolean}
 */
var validateStackSequences = function (pushed, popped) {
    let s = [];
    let q = 0;
    for (let num of pushed) {
        s.push(num);
        while (s.length > 0 && s[s.length - 1] == popped[q]) {
            ++q;
            s.pop();
        }
    }
    return s.length == 0;
};

C++

class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
        stack<int> s;
        int i = 0;
        for (int x : pushed) {
            s.push(x);
            while (!s.empty() && s.top() == popped[i]) {
                s.pop();
                ++i;
            }
        }
        return s.empty();
    }
};

TypeScript

function validateStackSequences(pushed: number[], popped: number[]): boolean {
    const stack = [];
    let i = 0;
    for (const num of pushed) {
        stack.push(num);
        while (stack.length !== 0 && stack[stack.length - 1] === popped[i]) {
            stack.pop();
            i++;
        }
    }
    return stack.length === 0;
}

Rust

impl Solution {
    pub fn validate_stack_sequences(pushed: Vec<i32>, popped: Vec<i32>) -> bool {
        let mut stack = Vec::new();
        let mut i = 0;
        for &num in pushed.iter() {
            stack.push(num);
            while !stack.is_empty() && *stack.last().unwrap() == popped[i] {
                stack.pop();
                i += 1;
            }
        }
        stack.len() == 0
    }
}

C#

public class Solution {
    public bool ValidateStackSequences(int[] pushed, int[] popped) {
        Stack<int> ans = new Stack<int>();
        int q = 0;
        foreach (int x in pushed)
        {
            ans.Push(pushed[x]);
            while (ans.Count != 0 && ans.Peek() == popped[q]) {
                ans.Pop();
                q += 1;
            }
        }
        return ans.Count == 0;
    }
}

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