161. One-Edit-Distance

difficulty: Medium

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.

Method One

class Solution {
    public boolean isOneEditDistance(String s, String t) {
        int m = s.length();
        int n = t.length();
        if( n < m ) {
            return isOneEditDistance(t, s);
        }
        // 设定 m <= n;
        // 如果长度差大于 1, 无解
        if( n - m > 1) {
            return false;
        }
        
        //至多有一个位置不一样。
        // 举例子, 两种情况，第一种长度相等。
        // abbb
        // acbb 
        // 如果某位不同，那么直接比较后面的子字符串
        // 第二种长度只差一位
        // abbb
        // acbbb
        // 如果某位不同，那么我们看看去掉长的字符串的这位之后，两个子数组是否还一致
        for(int i = 0; i < m; i++ ) {
            if ( s.charAt(i) == t.charAt(i) ) {
                continue;
            }
            if( n == m ) {
                return s.substring(i + 1, m).compareTo( t.substring( i + 1, n) ) == 0;
            }else{
                return s.substring(i, m).compareTo( t.substring(i + 1, n) ) == 0;
            }
        }
        return !(m == n);    // 排除两者完全相等的情况。
        
    }
}
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