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

题目地址：One Edit Distance - LeetCode

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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 - 优快云博客

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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.

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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,s
            lenS,lenT=lenT,lenS
        
        if lenS==lenT:
            count=0
            for i in range(0,lenS):
                if s[i]!=t[i]:
                    count+=1
                    if count>=2:
                        break
            if count==1:
                return True
            else:
                return False
        elif lenT-lenS>=2:
            return False
        else:
            flag=0
            for i in range(0,lenT):
                if i==lenT-1 and flag==0:
                    return True
                elif s[i-flag]!=t[i]:
                    if flag==0:
                        flag=1
                    else:
                        return False
            return 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 length
        if (ns == nt)
          return s.substring(i + 1).equals(t.substring(i + 1));
        // if strings have different lengths
        else
          return 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;
    }
};