POJ-1936 All in All

POJ-1936 All in All

题目链接：POJ-1936

题目大意：问前面的字符串是否按顺序存在于后面的字符串中

解题思路：双指针做法 十足的水题

代码块：

#include<iostream>
#include<string>

using namespace std;

int main(){
	string strA, strB;
	while(cin>>strA>>strB){
		int lenA = strA.length();
		int lenB = strB.length();
		int i = 0, j = 0;
		while(i < lenA && j < lenB){
			if(strA[i] == strB[j]){
				i++;
			}
			j++;
		}
		if(i == lenA) cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}
	return 0;
}