第31题:LeetCode946. Validate Stack Sequences验证栈的序列

题目

给定 pushed 和 popped 两个序列,只有当它们可能是在最初空栈上进行的推入 push 和弹出 pop 操作序列的结果时,返回 true;否则,返回 false 。

 

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。

考点

1.stack

2.vector


思路

输入:pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
输出:true

 

输入:pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
输出:false

 

函数入口做鲁棒性测试,都为空 ->true; size不同->false

flag=false;

定义两个iterator访问pushed和popped容器,用于读取元素,并且判断匹配成功标准

auto pushIt=pushed.begin();

大循环(popped没有访问完时)

{

入栈操作: data为空或者data.top!=*popIt时:

1.1 有数可以入:pushIt!=pushed.end() :

{

       data.push(*pushIt);

       pushIt++;

}

1.2 否则无数可入: break;

}

否则出栈:

{

data.pop();

popIt++:

}

}

匹配成功条件:stack为空,且popIt访问到了popped.end()


代码

newcoder

class Solution {
public:
    bool IsPopOrder(vector<int> pushV,vector<int> popV) {
        bool flag = false;
        
        if(!pushV.size()||!popV.size())
            return flag;
        
        //定义栈,读取迭代器 
        stack<int> data;
        auto  pushNext=pushV.begin();
        auto popNext=popV.begin(); 
        
        while(popNext!=popV.end())
        { //入栈
            if(data.empty()||data.top()!=*popNext)
                {
                    //pushed没有可入栈元素,返回
                    if(pushNext==pushV.end())
                        break;
                    else//否则,压入栈中
                    {
                        data.push(*pushNext);
                        pushNext++;
                    }
                }
            else//pop
            {
                data.pop();
                popNext++;//访问下个poped元素
            }
            
        }
        
        
        //如果压入栈为空,且poped序列访问完,则返回真
        if(data.empty()&&popNext==popV.end())
            flag=true;
        
        
        return flag;
    }
};

leetcode 

class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {   
        if(pushed.empty()&&popped.empty())
            return true;
        
        if(pushed.size()!=popped.size())
            return false;
        
        bool flag=false;
        
        //局部变量
        auto pushIt=pushed.begin();
        auto popIt=popped.begin();
        stack<int> data;
        
        
        while(popIt!=popped.end())
        {
                //进栈
            if(data.empty()||data.top()!=*popIt)
            {
                //无数可进
                if(pushIt==pushed.end())
                    break;
                else
                {
                    //进栈,pushIt++
                    data.push(*pushIt);
                    pushIt++;
                }
            }
            else
                //出栈
            {
                data.pop();
                popIt++;
            }
        }
        
        return (popIt==popped.end()&& data.empty()) ? true : flag ;
        
            
         
    }
};

问题

1. vector::begin/end

#include <iostream>
#include <vector>

int main ()
{
  std::vector<int> myvector;
  for (int i=1; i<=5; i++) myvector.push_back(i);

  std::cout << "myvector contains:";
  for (std::vector<int>::iterator it = myvector.begin() ; it != myvector.end(); ++it)
    std::cout << ' ' << *it;
  std::cout << '\n';

  return 0;
}
myvector contains: 1 2 3 4 5

用迭代器就不用知道序列的长度了~

容器的迭代器和数组的指针是一个作用。

posted @ 2019-02-08 21:37 lightmare 阅读(...) 评论(...) 编辑 收藏