剑指offer(16)栈的压入、弹出序列

题目:

  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可能为该栈的弹出顺序。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)

思路:

  添加一个辅助栈

  我们将用上面一个指针,pNextPush,进行移动,每移动一次将里面的数压栈,直到栈顶与下面一个指针指向的数相同时停下,

(至于为什么pNextPush跑到5,详情见代码)

  接下来我们弹栈,把下面一个指针pNextPop向前移动,发现不相等,那么我们就执行上面一步

以此类推,最后当我们得到pNextPop走到最后,而且栈为空时,我们就可以得出题设成立,接下来贴出一段书中不成立的例子,就不具体分析了

  

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import java.util.ArrayList;
import java.util.Stack;
public class Solution {
    public boolean IsPopOrder(int [] pushA,int [] popA) {
     boolean bPossible = false;
        if(pushA.length==0&&popA.length==0)
            return bPossible;
         
            int pNextPush = 0;
            int pNextPop = 0;
            Stack stack = new Stack();
             
            while(pNextPop<pushA.length){
                while(stack.isEmpty()||(int)stack.peek()!=popA[pNextPop]){
                   
                    if(pNextPush==pushA.length)
                        break;
                    
                    stack.push(pushA[pNextPush]);
                    pNextPush++;
                }
                 
                if((int)stack.peek()!=popA[pNextPop])
                    break;
                if(pNextPop<popA.length&&(int)stack.peek()==popA[pNextPop]){
                    int out = (int)stack.pop();
                    System.out.println(out);
                    pNextPop++;
                }
                 
                
            }
            if(stack.isEmpty()&&pNextPop==popA.length){
                bPossible = true;
            }
        
        return bPossible;
    }
}

  这道题的判断条件非常值得推敲,第一个while:

   while(pNextPop<pushA.length)如果这里把它改成pNextPush<Length,那么我们的循环将会提前结束,也就是如下图情况

 

  接下来还有中间的判断

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public boolean IsPopOrder(int [] pushA,int [] popA) {
    boolean bPossible = false;
    if(pushA.length==0&&popA.length==0)
        return bPossible;
     
        int pNextPush = 0;
        int pNextPop = 0;
        Stack stack = new Stack();
         
        while(pNextPop<pushA.length){
            while(stack.isEmpty()||(int)stack.peek()!=popA[pNextPop]){
                
    if(pNextPush==pushA.length)
                    break;
                 
                stack.push(pushA[pNextPush]);
                pNextPush++;
            }
            if(pNextPop<popA.length&&(int)stack.peek()==popA[pNextPop]){
                int out = (int)stack.pop();
                pNextPop++;
            }
            /*这是我第一版的代码
        这里这么写主要是为了防止死循环,重新进入循环内第一个while
        然而这也导致了卡在了上图的时候,就直接跳出循环*/
 
    if(pNextPush==pushA.length)
        break;
            
        }
        if(stack.isEmpty()&&pNextPop==popA.length){
            bPossible = true;
        }
    
    return bPossible;
}                                                                         

  

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while(pNextPush<pushA.length){
                while(stack.isEmpty()||(int)stack.peek()!=popA[pNextPop]){
                  
            if(pNextPush==pushA.length)
                        break;
       
                    stack.push(pushA[pNextPush]);
                    pNextPush++;
                }
                 
                 
                if(pNextPop<popA.length&&(int)stack.peek()==popA[pNextPop]){
                    int out = (int)stack.pop();
                    System.out.println(out);
                    pNextPop++;
                }
/*这是第二版的循环体,显然不科学,当4弹出时,5还未进栈,条件成立,直接退出*/
        if((int)stack.peek()!=popA[pNextPop])
                    break;
                 
                
            }

  

 

posted @   FigSprite  阅读(159)  评论(0编辑  收藏  举报
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