leetcode 161: One Edit Distance

标签：

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

[分析]

delete, add, replace 三种都是 One edit distance. delete 和add 两个string长度差1, replace相同长度.

[注意]

None

[CODE]

public class Solution {
    public boolean isOneEditDistance(String s, String t) {
        if(s==null || t==null) return false;
        
        if(Math.abs(s.length() - t.length())>1) {
            return false;
        }
        
        if(s.length() == t.length()) {
            int count = 0;
            for(int i=0; i<s.length(); i++) {
                if(s.charAt(i) != t.charAt(i)) ++count;
                if(count > 1) return false;
            }
            return count==1;
        } else {
            String ss,ls;
            if(s.length() < t.length() ) {
                ss = s;
                ls = t;
            } else {
                ss = t;
                ls = s;
            }
            int i=0, j=0;
            int count = 0;
            while(i<ss.length() && j<ls.length()) {
                if(ss.charAt(i) != ls.charAt(j) ) {
                    ++count; ++j; 
                    if(count > 1) return false;
                } else {
                    ++i; ++j;
                }
            }
            return true;
        }
    }
}

leetcode 161: One Edit Distance

标签：

原文地址：http://blog.csdn.net/xudli/article/details/42340867