题解 | #栈的压入、弹出序列#

需要同时遍历两个数组的全部元素，时间复杂度O(n); 借助一个栈，最坏的情况下，全部入栈，空间复杂度O(n)。

import java.util.Stack;

public class Solution { public boolean IsPopOrder(int [] pushA,int [] popA) { if(pushA.length != popA.length) return false; Stack stack = new Stack<>(); stack.push(pushA[0]); int i, j; for(i = 1,j = 0; i < pushA.length && j < popA.length; ){ if( stack.isEmpty() !=true && popA[j] == stack.peek()){ stack.pop(); j++; }else{ stack.push(pushA[i]); i++; } } while(j < popA.length){ if(popA[j] == stack.peek()){ stack.pop(); j++; }else return false; } return true; } }