Stack [] [] to prove safety Offer stack push, pop sequence

Title: two input sequence of integers, the first sequence indicating whether a sequence of stack of press-fitting, the second sequence is determined may make the pop-up sequence for the stack. All figures are not pushed onto the stack is assumed equal. Such as a sequence of a sequence of 1,2,3,4,5 is pressed into the stack, the push sequence is 4,5,3,2,1 sequence corresponding to a sequence of pop, but 4,3,5,1,2 it is impossible to push pop-up sequence of the sequence. (Note: the length of the two sequences are equal)

 

A: using an auxiliary stack, the first sequence numbers are sequentially pushed into the auxiliary stack, then the top element and the comparison loop stack sequences are equal

　　If the last cycle, the auxiliary stack is empty, return true

　　False otherwise

 

class Solution {
public:
    bool IsPopOrder(vector<int> pushV,vector<int> popV) {
        if(pushV.empty()|| pushV.size() != popV.size())
        {
            return false;
        }
        stack<int> m_stack;
        
        for(int i = 0,j = 0; i < pushV.size(); i++)
        {
            m_stack.push(pushV[i]);
            while(!m_stack.empty() && (popV[j] == m_stack.top()))
            {
                m_stack.pop();
                j++;
            }
        }
        if(m_stack.empty())
        {
            return true;
        }
        return false;
    }
};

　　

 

 

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