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legolas007
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package com.usher.algorithm.offer;
import java.util.Stack;
/**
* @Author: Usher
* @Description:
* 栈的压入,弹出序列
* 链接:https://www.nowcoder.com/questionTerminal/d77d11405cc7470d82554cb392585106
来源:牛客网
借用一个辅助的栈,遍历压栈顺序,先讲第一个放入栈中,这里是1,然后判断栈顶元素是不是出栈顺序的第一个元素,这里是4,很显然1≠4,所以我们继续压栈,直到相等以后开始出栈,出栈一个元素,则将出栈顺序向后移动一位,直到不相等,这样循环等压栈顺序遍历完成,如果辅助栈还不为空,说明弹出序列不是该栈的弹出顺序。
举例:
入栈1,2,3,4,5
出栈4,5,3,2,1
首先1入辅助栈,此时栈顶1≠4,继续入栈2
此时栈顶2≠4,继续入栈3
此时栈顶3≠4,继续入栈4
此时栈顶4=4,出栈4,弹出序列向后一位,此时为5,,辅助栈里面是1,2,3
此时栈顶3≠5,继续入栈5
此时栈顶5=5,出栈5,弹出序列向后一位,此时为3,,辅助栈里面是1,2,3
….
依次执行,最后辅助栈为空。如果不为空说明弹出序列不是该栈的弹出顺序。
*/
public class Solution31 {
public boolean IsPopOrder(int [] pushA,int [] popA) {
Stack<Integer> stack = new Stack<>();
//用于标识弹出序列的位置
int popIndex = 0;
for (int i =0;i < pushA.length;i++){
stack.push(pushA[i]);
//如果栈不为空,且栈顶元素等于弹出序列
while (!stack.isEmpty() && stack.peek() == popA[popIndex]){
stack.pop();
//弹出序列向后一位
popIndex++;
}
}
return stack.isEmpty();
}
}
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