[Thoughts]
这是个简化版。只要考虑几种可能, 如果s.length == t.length,只要看看不相同的字符是否超过一个即可。如果s.length < t.length,看看在s中插入一个字符是否能解决问题。
[Code]
1: bool isOneEditDistance(string s, string t) {
2: int lena = s.size();
3: int lenb = t.size();
4: if(lenb > lena) return isOneEditDistance(t, s);
5: if(lena - lenb > 1) return false;
6:
7: int diff = 0;
8: for(int i =0, j = 0; i< s.size() && j< t.size() && diff < 2; i++, j++) {
9: if(s[i] != t[j]) {
10: diff++;
11: if(lena != lenb) t.insert(j, 1, s[i]);
12: }
13: }
14:
15: if((lena != lenb) && diff <2) return true;
16: if((lena == lenb) && diff ==1) return true;
17: return false;
18: }
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