[leetcode]161. One Edit Distance编辑步数为一

Given two strings s and t, determine if they are both one edit distance apart.

Note: 

There are 3 possiblities to satisify one edit distance apart:

1. Insert a character into s to get t
2. Delete a character from s to get t
3. Replace a character of s to get t

Example 1:

Input: s = "ab", t = "acb"
Output: true
Explanation: We can insert 'c' into s to get t.

Example 2:

Input: s = "cab", t = "ad"
Output: false
Explanation: We cannot get t from s by only one step.

Example 3:

Input: s = "1203", t = "1213"
Output: true
Explanation: We can replace '0' with '1' to get t.

题目

给定两个字符串，判断其编辑步数是否为1

思路

此题可算是[leetcode]72. Edit Distance 最少编辑步数的一个拆分简化版本

代码

class Solution {
     public boolean isOneEditDistance(String s, String t) {
        int m = s.length(), n = t.length();
        if(m == n) return isOneModified(s, t);
        if(m - n == 1) return isOneDeleted(s, t);
        if(n - m == 1) return isOneDeleted(t, s);
        // 长度差距大于2直接返回false
        return false;
    }
    
    private boolean isOneModified(String s, String t){
        boolean modified = false;
        // 看是否只修改了一个字符
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i) != t.charAt(i)){
                if(modified) return false;
                modified = true;
            }
        }
        return modified;
    }
    
    public boolean isOneDeleted(String longer, String shorter){
        // 找到第一组不一样的字符，看后面是否一样
        for(int i = 0; i < shorter.length(); i++){
            if(longer.charAt(i) != shorter.charAt(i)){
                return longer.substring(i + 1).equals(shorter.substring(i));
            }
        }
        return true;
    }
}