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feat: add solutions to lc problem: No.0161
No.0161.One Edit Distance
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‎solution/0100-0199/0161.One Edit Distance/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：分情况讨论**
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记 $m$ 表示字符串 $s$ 的长度，$n$ 表示字符串 $t$ 的长度。我们可以假定 $m$ 恒大于等于 $n$。
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若 $m-n\gt1$，直接返回 false；
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否则，遍历 $s$ 和 $t$，若遇到 $s[i]$ 不等于 $t[i]$：
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- 若 $m \neq n$，比较 $s[i+1:]$ 与 $t[i:]$，相等返回 true，否则返回 false；
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- 若 $m = n$，比较 $s[i:]$ 与 $t[i:]$，相等返回 true，否则返回 false。
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遍历结束，说明遍历过的 $s$ 跟 $t$ 所有字符相等，此时需要满足 $m=n+1$。
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时间复杂度 $O(m)$，空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def isOneEditDistance(self, s: str, t: str) -> bool:
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if len(s) < len(t):
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return self.isOneEditDistance(t, s)
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m, n = len(s), len(t)
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if m - n > 1:
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return False
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for i, c in enumerate(t):
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if c != s[i]:
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return s[i + 1:] == t[i + 1:] if m == n else s[i + 1:] == t[i:]
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return m == n + 1
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean isOneEditDistance(String s, String t) {
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int m = s.length(), n = t.length();
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if (m < n) {
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return isOneEditDistance(t, s);
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}
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if (m - n > 1) {
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return false;
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}
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for (int i = 0; i < n; ++i) {
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if (s.charAt(i) != t.charAt(i)) {
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if (m == n) {
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return s.substring(i + 1).equals(t.substring(i + 1));
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}
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return s.substring(i + 1).equals(t.substring(i));
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}
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}
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return m == n + 1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool isOneEditDistance(string s, string t) {
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int m = s.size(), n = t.size();
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if (m < n) return isOneEditDistance(t, s);
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if (m - n > 1) return false;
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for (int i = 0; i < n; ++i) {
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if (s[i] != t[i]) {
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if (m == n) return s.substr(i + 1) == t.substr(i + 1);
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return s.substr(i + 1) == t.substr(i);
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}
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}
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return m == n + 1;
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}
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};
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```
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### **Go**
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```go
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func isOneEditDistance(s string, t string) bool {
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m, n := len(s), len(t)
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if m < n {
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return isOneEditDistance(t, s)
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}
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if m-n > 1 {
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return false
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}
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for i := range t {
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if s[i] != t[i] {
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if m == n {
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return s[i+1:] == t[i+1:]
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}
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return s[i+1:] == t[i:]
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}
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}
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return m == n+1
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}
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```
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### **...**

‎solution/0100-0199/0161.One Edit Distance/README_EN.md

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### **Python3**
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```python
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class Solution:
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def isOneEditDistance(self, s: str, t: str) -> bool:
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if len(s) < len(t):
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return self.isOneEditDistance(t, s)
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m, n = len(s), len(t)
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if m - n > 1:
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return False
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for i, c in enumerate(t):
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if c != s[i]:
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return s[i + 1:] == t[i + 1:] if m == n else s[i + 1:] == t[i:]
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return m == n + 1
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```
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### **Java**
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```java
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class Solution {
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public boolean isOneEditDistance(String s, String t) {
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int m = s.length(), n = t.length();
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if (m < n) {
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return isOneEditDistance(t, s);
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}
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if (m - n > 1) {
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return false;
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}
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for (int i = 0; i < n; ++i) {
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if (s.charAt(i) != t.charAt(i)) {
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if (m == n) {
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return s.substring(i + 1).equals(t.substring(i + 1));
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}
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return s.substring(i + 1).equals(t.substring(i));
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}
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}
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return m == n + 1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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bool isOneEditDistance(string s, string t) {
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int m = s.size(), n = t.size();
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if (m < n) return isOneEditDistance(t, s);
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if (m - n > 1) return false;
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for (int i = 0; i < n; ++i) {
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if (s[i] != t[i]) {
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if (m == n) return s.substr(i + 1) == t.substr(i + 1);
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return s.substr(i + 1) == t.substr(i);
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}
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}
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return m == n + 1;
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}
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};
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```
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### **Go**
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```go
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func isOneEditDistance(s string, t string) bool {
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m, n := len(s), len(t)
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if m < n {
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return isOneEditDistance(t, s)
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}
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if m-n > 1 {
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return false
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}
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for i := range t {
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if s[i] != t[i] {
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if m == n {
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return s[i+1:] == t[i+1:]
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}
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return s[i+1:] == t[i:]
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}
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}
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return m == n+1
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}
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```
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### **...**

‎solution/0100-0199/0161.One Edit Distance/Solution.cpp

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

‎solution/0100-0199/0161.One Edit Distance/Solution.go

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func isOneEditDistance(s string, t string) bool {
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m, n := len(s), len(t)
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if m < n {
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return isOneEditDistance(t, s)
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}
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if m-n > 1 {
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return false
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}
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for i := range t {
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if s[i] != t[i] {
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if m == n {
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return s[i+1:] == t[i+1:]
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}
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return s[i+1:] == t[i:]
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}
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}
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return m == n+1
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}

‎solution/0100-0199/0161.One Edit Distance/Solution.java

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class Solution {
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public boolean isOneEditDistance(String s, String t) {
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int m = s.length(), n = t.length();
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if (m < n) {
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return isOneEditDistance(t, s);
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}
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if (m - n > 1) {
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return false;
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}
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for (int i = 0; i < n; ++i) {
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if (s.charAt(i) != t.charAt(i)) {
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if (m == n) {
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return s.substring(i + 1).equals(t.substring(i + 1));
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}
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return s.substring(i + 1).equals(t.substring(i));
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}
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}
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return m == n + 1;
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}
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}

‎solution/0100-0199/0161.One Edit Distance/Solution.py

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class Solution:
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def isOneEditDistance(self, s: str, t: str) -> bool:
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if len(s) < len(t):
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return self.isOneEditDistance(t, s)
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m, n = len(s), len(t)
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if m - n > 1:
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return False
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for i, c in enumerate(t):
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if c != s[i]:
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return s[i + 1:] == t[i + 1:] if m == n else s[i + 1:] == t[i:]
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return m == n + 1

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