LeetCode - One Edit Distance
2015-04-07 00:42
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https://leetcode.com/problems/one-edit-distance/
Given two strings S and T, determine if they are both one edit distance apart.
One Edit Distance即是:
1. 插入一个字符
2. 更改一个字符, 如ab->ac
所以分情况是:
1. 如果两个字符串等长,那么其中有一个字符不相同就可以
2. 如果两个字符串长度相差1,那么除了被插入的字符,其余字符相同
3. 如果两个字符串长度相差大于1,那么肯定不只1个edit distance了
看了别人更简洁的写法:
合并两种情况!
Given two strings S and T, determine if they are both one edit distance apart.
One Edit Distance即是:
1. 插入一个字符
2. 更改一个字符, 如ab->ac
所以分情况是:
1. 如果两个字符串等长,那么其中有一个字符不相同就可以
2. 如果两个字符串长度相差1,那么除了被插入的字符,其余字符相同
3. 如果两个字符串长度相差大于1,那么肯定不只1个edit distance了
public class Solution { public boolean isOneEditDistance(String s, String t) { int m = s.length(); int n = t.length(); if(m>n) return isOneEditDistance(t, s); int shift = n-m; if(shift>1) return false; if(shift==0){ int i = 0; while(i<m && s.charAt(i)==t.charAt(i)) i++; if(i==m) return false; i++; while(i<m && s.charAt(i) == t.charAt(i)) i++; return i==m; } else{ int i = 0; while(i<m && s.charAt(i)==t.charAt(i)) i++; if(i==m) return true; while(i<m && s.charAt(i)==t.charAt(i+shift)) i++; return i==m; } } }
看了别人更简洁的写法:
public class Solution { public boolean isOneEditDistance(String s, String t) { int m = s.length(); int n = t.length(); if(m>n) return isOneEditDistance(t, s); int shift = n-m; if(shift>1) return false; int i = 0; while(i<m && s.charAt(i)==t.charAt(i)) i++; if(i==m) return shift>0; if(shift==0) i++; while(i<m && s.charAt(i)==t.charAt(i+shift)) i++; return i==m; } }
合并两种情况!
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