LeetCode 161. One Edit Distance--Python,Java,C++解法

题目地址：One Edit Distance - LeetCode

此题链接：One Edit Distance - LeetCode
这道题目进阶版本Edit Distance - LeetCode
文章地址:LeetCode 72. Edit Distance - zhangpeterx的博客 - CSDN博客

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.

如果s与t是One Edit Distance，那么s与t要么长度相等，做一次替换，要么长度差一，做删除或者插入。
Python解法如下：

class Solution:def isOneEditDistance(self, s: str, t: str) -> bool:lenS=len(s)lenT=len(t)if lenS>lenT:s,t=t,slenS,lenT=lenT,lenSif lenS==lenT:count=0for i in range(0,lenS):if s[i]!=t[i]:count+=1if count>=2:breakif count==1:return Trueelse:return Falseelif lenT-lenS>=2:return Falseelse:flag=0for i in range(0,lenT):if i==lenT-1 and flag==0:return Trueelif s[i-flag]!=t[i]:if flag==0:flag=1else:return Falsereturn True

Java解法如下：

class Solution {public boolean isOneEditDistance(String s, String t) {int ns = s.length();int nt = t.length();// Ensure that s is shorter than t.if (ns > nt)return isOneEditDistance(t, s);// The strings are NOT one edit away distance  // if the length diff is more than 1.if (nt - ns > 1)return false;for (int i = 0; i < ns; i++)if (s.charAt(i) != t.charAt(i))// if strings have the same lengthif (ns == nt)return s.substring(i + 1).equals(t.substring(i + 1));// if strings have different lengthselsereturn s.substring(i).equals(t.substring(i + 1));// If there is no diffs on ns distance// the strings are one edit away only if// t has one more character. return (ns + 1 == nt);}
}

C++解法如下：

class Solution {public:bool isOneEditDistance(string s, string t) {int m = s.size(), n = t.size();if (m > n) {return isOneEditDistance(t, s);}for (int i = 0; i < m; i++) {if (s[i] != t[i]) {if (m == n) {return s.substr(i + 1) == t.substr(i + 1);}return s.substr(i) == t.substr(i + 1);}}return m + 1 == n;}
};

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