feat: add python and java solutions to lcci question

添加《程序员面试金典》题解：面试题 01.05. 一次编辑

Author: yanglbme

lcci/01.05.One Away/README.md

@@ -26,20 +26,53 @@ second = "pal"
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+遍历两个字符串，逐个字符比较判断。
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def oneEditAway(self, first: str, second: str) -> bool:
+        n1, n2 = len(first), len(second)
+        if abs(n1 - n2) > 1:
+            return False
+        if n1 + n2 <= 2:
+            return True
+        if first[0] != second[0]:
+            if n1 == n2:
+                return first[1:] == second[1:]
+            else:
+                return first[1:] == second or second[1:] == first
+        else:
+            return self.oneEditAway(first[1:], second[1:])
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public boolean oneEditAway(String first, String second) {
+        int n1 = first.length(), n2 = second.length();
+        int differ = Math.abs(n1 - n2);
+        if (differ > 1) {
+            return false;
+        }
+        if (n1 + n2 <= 2) {
+            return true;
+        }
+        if (first.charAt(0) != second.charAt(0)) {
+            if (n1 == n2) {
+                return first.substring(1).equals(second.substring(1));
+            } else {
+                return first.substring(1).equals(second) || second.substring(1).equals(first);
+            }
+        } else {
+            return oneEditAway(first.substring(1), second.substring(1));
+        }
+    }
+}
 ```
 
 ### ...

lcci/01.05.One Away/README_EN.md

@@ -1,46 +1,100 @@
-# [01.05. One Away](https://leetcode-cn.com/problems/one-away-lcci)
-
-## Description
-<p>There are three types of edits that can be performed on strings: insert a character, remove a character, or replace a character. Given two strings, write a function to check if they are one edit (or zero edits) away.</p>
-
-<p>&nbsp;</p>
-
-<p><strong>Example&nbsp;1:</strong></p>
-
-<pre>
-<strong>Input:</strong> 
-first = &quot;pale&quot;
-second = &quot;ple&quot;
-<strong>Output:</strong> True
-</pre>
-
-<p><strong>Example&nbsp;2:</strong></p>
-
-<pre>
-<strong>Input:</strong> 
-first = &quot;pales&quot;
-second = &quot;pal&quot;
-<strong>Output:</strong> False
-</pre>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [01.05. One Away](https://leetcode-cn.com/problems/one-away-lcci)
+
+## Description
+<p>There are three types of edits that can be performed on strings: insert a character, remove a character, or replace a character. Given two strings, write a function to check if they are one edit (or zero edits) away.</p>
+
+
+
+<p>&nbsp;</p>
+
+
+
+<p><strong>Example&nbsp;1:</strong></p>
+
+
+
+<pre>
+
+<strong>Input:</strong> 
+
+first = &quot;pale&quot;
+
+second = &quot;ple&quot;
+
+<strong>Output:</strong> True
+
+</pre>
+
+
+
+<p><strong>Example&nbsp;2:</strong></p>
+
+
+
+<pre>
+
+<strong>Input:</strong> 
+
+first = &quot;pales&quot;
+
+second = &quot;pal&quot;
+
+<strong>Output:</strong> False
+
+</pre>
+
+
+
+
+## Solutions
+
+
+### Python3
+
+```python
+class Solution:
+    def oneEditAway(self, first: str, second: str) -> bool:
+        n1, n2 = len(first), len(second)
+        if abs(n1 - n2) > 1:
+            return False
+        if n1 + n2 <= 2:
+            return True
+        if first[0] != second[0]:
+            if n1 == n2:
+                return first[1:] == second[1:]
+            else:
+                return first[1:] == second or second[1:] == first
+        else:
+            return self.oneEditAway(first[1:], second[1:])
+```
+
+### Java
+
+```java
+class Solution {
+    public boolean oneEditAway(String first, String second) {
+        int n1 = first.length(), n2 = second.length();
+        int differ = Math.abs(n1 - n2);
+        if (differ > 1) {
+            return false;
+        }
+        if (n1 + n2 <= 2) {
+            return true;
+        }
+        if (first.charAt(0) != second.charAt(0)) {
+            if (n1 == n2) {
+                return first.substring(1).equals(second.substring(1));
+            } else {
+                return first.substring(1).equals(second) || second.substring(1).equals(first);
+            }
+        } else {
+            return oneEditAway(first.substring(1), second.substring(1));
+        }
+    }
+}
+```
+
+### ...
+```
+
+```

lcci/01.05.One Away/Solution.java

@@ -0,0 +1,21 @@
+class Solution {
+    public boolean oneEditAway(String first, String second) {
+        int n1 = first.length(), n2 = second.length();
+        int differ = Math.abs(n1 - n2);
+        if (differ > 1) {
+            return false;
+        }
+        if (n1 + n2 <= 2) {
+            return true;
+        }
+        if (first.charAt(0) != second.charAt(0)) {
+            if (n1 == n2) {
+                return first.substring(1).equals(second.substring(1));
+            } else {
+                return first.substring(1).equals(second) || second.substring(1).equals(first);
+            }
+        } else {
+            return oneEditAway(first.substring(1), second.substring(1));
+        }
+    }
+}

lcci/01.05.One Away/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def oneEditAway(self, first: str, second: str) -> bool:
+        n1, n2 = len(first), len(second)
+        if abs(n1 - n2) > 1:
+            return False
+        if n1 + n2 <= 2:
+            return True
+        if first[0] != second[0]:
+            if n1 == n2:
+                return first[1:] == second[1:]
+            else:
+                return first[1:] == second or second[1:] == first
+        else:
+            return self.oneEditAway(first[1:], second[1:])