基础题:栈
1:输入一个压栈序列,判断第二个序列是否为其出栈序列。
例如:入栈序列:1 2 3 4 5 6,出栈序列,4,3,5,2,6,1
算法思想:
- 1:根据出栈序列,入栈,直到其栈顶等于出栈元素,栈s:4,3,2,1
- 2:栈顶与出栈序列相同出栈,否则break
根据入栈序列入栈:(左为栈顶)
栈:1 2 3 4 1 2 3 1 2 5 1 2 1 6 1 |空
出栈元素: 4 3 5 2 6 1 , 4 3 5 2 6 1 ,4 35 2 6 1,4 3 5 2 6 1 ,4 3 5 26 1 ,4 3 5 2 61 ,完
示例代码:
#include <iostream> #include <stack> using namespace std; bool IsStackPopOrder(int *pushorder,int *poporder,int len) { bool isorder = false; if(pushorder!=NULL && poporder != NULL && len > 0) { stack<int> s; int *pnextpush = pushorder; int *pnextpop = poporder; while((pnextpop - poporder) < len) { while(s.empty()||s.top()!=*pnextpop) { if((pnextpush - pushorder)==len)break; s.push(*pnextpush); pnextpush++; } if (s.top() == *pnextpop) { s.pop(); pnextpop++; } else break; } if ((pnextpop - poporder)==len && s.empty())isorder = true; } return isorder; } void main() { int array1[7] = {1,2,3,4,5,6,7}; int array2[7] = {4,3,5,6,7,1,2}; if(IsStackPopOrder(array1,array2,7))cout<<"是"; else cout<<"否"; system("pause"); }
参考:《剑指offer》