E - Substring Reverse

最新推荐文章于 2023-11-15 11:44:27 发布

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最新推荐文章于 2023-11-15 11:44:27 发布

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分类专栏： 2020寒假比赛题目

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本文介绍了一种算法，用于判断是否能通过反转一个给定字符串的某个子串使其与目标字符串相等。通过前后查找不匹配字符并比较反转后的子串，实现高效匹配。

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Two strings s and t of the same length are given. Determine whether it is possible to make t from s using exactly one reverse of some its substring.

Input
The first line contains the string s, and the second — the string t. Both strings have the same length from 1 to 200000 characters and consist of lowercase Latin letters.

Output
Output «YES», if it is possible to reverse some substring of s to make s equal to t, and «NO», otherwise.

Examples
Input
abcdefg
abedcfg
Output
YES
Input
abcdefg
abdecfg
Output
NO
思路：从前开始找到不一样字母的下标，再从后找到不一样字母的下标。

#include<bits/stdc++.h>
using namespace std;
const int maxx=2e5+10;
char a[maxx],b[maxx];
int main()
{
    std::ios::sync_with_stdio(false);
    cin>>a>>b;
    int len=strlen(a);
    int n1=-1,n2=-1;
    for(int i=0; i<len; i++)
    {
        if(a[i]!=b[i])
        {
            n1=i;
            break;
        }
    }
    for(int i=len-1; i>=0; i--)
    {
        if(a[i]!=b[i])
        {
            n2=i;
            break;
        }
    }
    int flag=1;
    for(int i=n1; i<=n2; i++)
    {
        if(a[i]!=b[n2+n1-i])
        {
            flag=0;
            // cout<<i<<"&&&&"<<j<<endl;
            break;
        }
    }
    if(flag==1)
        cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;

    return 0;
}