剑指offer-题22：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是某栈的压入顺序，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。（注意：这两个序列的长度是相等的）

实验平台：牛客网

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解决思路：

java:

import java.util.Stack;

public class Solution {
    public boolean IsPopOrder(int [] pushA,int [] popA) {
        boolean possible = false;
        if (pushA.length > 0 && popA.length > 0) {
            int pushIndex = 0;
            int popIndex = 0;
            Stack<Integer> dataStack = new Stack<Integer>();
            while (popIndex < popA.length) {
                while (dataStack.isEmpty() || dataStack.peek() != popA[popIndex]) {
                    // 将压栈序列压入一个栈中直到： 1.栈顶元素和弹出序列的索引位置相同
                    // 2.压栈序列遍历结束也没有找到和弹出序列的索引位置相同的值
                    if (pushIndex == pushA.length) {
                        break;
                    }
                    dataStack.push(new Integer(pushA[pushIndex]));
                    pushIndex++;
                }
                if (dataStack.peek() != popA[popIndex]) {
                    break;
                }
                dataStack.pop();
                popIndex++;
            }
            if(dataStack.isEmpty() && popIndex == popA.length){
                possible = true;
            }
        }
        return possible;
    }
}

python:

# -*- coding:utf-8 -*-
class Solution:
    def IsPopOrder(self, pushV, popV):
        # write code here
        possible = False
        if len(pushV) > 0 and len(popV) > 0:
            push_index = 0
            pop_index = 0
            stack = []
            while pop_index < len(popV):
                while len(stack) == 0 or stack[-1] != popV[pop_index]:
                    if push_index == len(pushV):
                        break
                    stack.append(pushV[push_index])
                    push_index += 1
                if stack[-1] != popV[pop_index]:
                    break
                stack.pop()
                pop_index += 1
            if pop_index == len(popV) and len(stack) == 0:
                possible = True
        return possible