161. One Edit Distance

Problem

Given two strings S and T, determine if they are both one edit distance apart.

Solution

只有一个编辑距离，那就是只能在增，删，减中的一步

Bug ： 两个字符串相等时应该返回 false

class Solution {
    
    bool helper( const string& s, int s_begin, const string& t, int t_begin){
        while(s_begin < s.size()){
            if(s[s_begin++] != t[t_begin++]) return false;
        }
        return true;
    }
public:
    bool isOneEditDistance(string s, string t) {
    
        const int s_len = s.size(), t_len = t.size();
        if( s_len < t_len ) return isOneEditDistance( t, s);
        if( s_len - t_len >= 2) return false;
        
        for( int i = 0; i < s_len ; i++){
            if(s[i] != t[i]) {
                int s_begin = i + 1, t_begin = ( s_len == t_len ? i + 1 : i );
                
                return helper(s, s_begin, t, t_begin);
            }
        }
        return false;
    }
};