[leetcode] 161. One Edit Distance @ python

一编辑距离

最新推荐文章于 2021-04-01 23:12:44 发布

原创最新推荐文章于 2021-04-01 23:12:44 发布 · 439 阅读

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本文探讨了判断两个字符串是否可以通过一次插入、删除或替换操作达到一编辑距离的方法。通过分类讨论并提供示例，展示了如何实现这一功能。

原题

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:

Insert a character into s to get t
Delete a character from s to get t
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.

解法

根据题意, s能转化为t的条件是两个的长度的差不超过1, 因此先排除那些长度的差大于1的情况. 然后分类讨论, 尝试将s变为t.

代码

class Solution(object):
    def isOneEditDistance(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        len_s, len_t = len(s), len(t)
        if not(len_s - len_t in [-1, 0, 1]):
            return False
        
        s, t = list(s), list(t)
        if len_s < len_t:
            # try to insert a char to get t
            for i, ch in enumerate(s):
                if s[i] != t[i]:
                    s.insert(i, t[i])
                    # check if they are equal
                    return s == t
            return True
        
        if len_s == len_t:
            for i, ch in enumerate(s):
                if s[i] != t[i]:
                    s[i] = t[i]
                    return s == t
        
        if len_s > len_t:
            # try to delete a char
            for i, ch in enumerate(t):
                if s[i] != t[i]:
                    s.pop(i)
                    return s == t
            return True