161. One-Edit-Distance.md

Description

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.

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分三种情况判断：

• s与t长度相等。看s和t是否有且只有一个字母不相同。
• s比t多一个字母 或 s比t少一个字母。这两种情况逻辑其实是一样的，同时遍历s和t，当遇到第一个不同的字母，看后面的部分是否一样。如果一样，删除这个字母后s=t
• 其他情况：s和t长度相差2以上，这样肯定不行。

python solution

class Solution(object):
    def isOneEditDistance(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        
        if len(s) == len(t):
            return sum(map(lambda c: s[c] != t[c], range(len(s)))) == 1

        while s and t and s[0] == t[0]: # 注意前面的s and t条件，保证s和t不能为空。
            s = s[1:]
            t = t[1:]

        if len(s) - len(t) == 1:
            return s[1:] == t
        elif len(t) - len(s) == 1:
            return t[1:] == s
        
        return False    # 长度相差2以上的，直接返回False