编程之美 3.1 字符串移位包含问题

  题目描述: 给定两个字符串s1, s2 , 要求判定s2是否能够被s1做循环移位得到的字符串包含

  解题思路: 复制一遍s1重新接到s1后面 , 再查询, O(nk)的时间复杂度, 还有一种不需要申请过多新空间的方法

  代码: 

#include <iostream>
#include <queue>
#include <string>
#include <vector>
#include <algorithm>
#include <list>
#include <iterator>
#include <cmath>
#include <cstring>
#include <forward_list>
using namespace std;


int main() {
    string s1,s2;
    cin >> s1 >> s2;
        s1 += s1;
    if(find(s1,s2)!=-1) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}    
View Code
#include <iostream>
#include <queue>
#include <string>
#include <vector>
#include <algorithm>
#include <list>
#include <iterator>
#include <cmath>
#include <cstring>
#include <forward_list>
using namespace std;

int my_find(string s1, string s2) {
    int len1 = (int)s1.length();
    int len2 = (int)s2.length();
    int found = 0;
    int cnt = 0;
    while(cnt < 2*len1) {
        int pos1 = cnt%len1;
        int pos2 = 0;
        if(s1[pos1] == s2[pos2]) {
            int i = 0;
            for(i = 0; i < len2; i++) {
                if(s1[(pos1+i)%len1] != s2[pos2+i]) break;
            }
            if(i == len2) {
                found = 1;
                break;
            }
        }
        cnt++;
    }
    if(found) return 1;
    else return 0;
}

int main() {
    string s1,s2;
    cin >> s1 >> s2;
    if(my_find(s1,s2)) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}
View Code

  思考: 都是脑洞题, 锻炼锻炼自己的思维也不错, 然后下面夯实自己的数据结构基础

posted on 2017-12-09 19:27  FriskyPuppy  阅读(192)  评论(0编辑  收藏  举报

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