剑指Offer 31 栈的压入、弹出序列

栈的压入、弹出序列

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）

# -*- coding:utf-8 -*-
class Solution:
    def IsPopOrder(self, pushV, popV):
        n = len(pushV)
        stack = []
        popout = [0] * n
        preidx = 0
        for i in range(len(pushV)):
            cur = popV[i]
            if pushV.count(cur) > 0:
                idx = pushV.index(cur)
                if idx >= preidx:
                    for j in range(preidx,idx):
                        if popout[j] == 1:
                            continue
                        else:
                            stack.append(pushV[j])
                else:
                    top = stack.pop(-1)
                    if top != cur:
                        return False
                popout[idx] = 1
                preidx = idx
            else:
                return False
        return True
        # write code here

 

Java版代码，leetcode地址：

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        int i = 0;//pushed数组的游标
        int m = pushed.length;
        int n = popped.length;
        if(n == 0) {
            return true;//空队列，直接返回true
        }        
        Stack<Integer> st = new Stack<Integer>();
        st.push(-1);
        for(int j=0;j<n;j++) {
            while(st.peek() != popped[j] && i<m) {
                st.push(pushed[i]);
                i++;
            }
            if(st.peek() == popped[j]) {
                st.pop();
            }else {
                return false;
            }
        }        
        return st.size() == 1 && st.peek() == -1;
    }
}