feat: add python and java solutions to leetcode problems

Author: yanglbme

README.md

@@ -112,8 +112,8 @@
 ### 数学
 
 - [只出现一次的数字](/solution/0100-0199/0136.Single%20Number/README.md)
-- [数组中数字出现的次数](/lcof/面试题56%20-%20I.%20数组中数字出现的次数/README.md)
-- [数组中数字出现的次数 II](/lcof/面试题56%20-%20II.%20数组中数字出现的次数%20II/README.md)
+- [只出现一次的数字 II](/solution/0100-0199/0137.Single%20Number%20II/README.md)
+- [只出现一次的数字 III](/solution/0200-0299/0260.Single%20Number%20III/README.md)
 - [错误的集合](/solution/0600-0699/0645.Set%20Mismatch/README.md)
 - [二进制中 1 的个数](/lcof/面试题15.%20二进制中1的个数/README.md)
 - [计数质数](/solution/0200-0299/0204.Count%20Primes/README.md)

README_EN.md

@@ -106,6 +106,8 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 ### Math
 
 - [Single Number](/solution/0100-0199/0136.Single%20Number/README_EN.md)
+- [Single Number II](/solution/0100-0199/0137.Single%20Number%20II/README_EN.md)
+- [Single Number III](/solution/0200-0299/0260.Single%20Number%20III/README_EN.md)
 - [Set Mismatch](/solution/0600-0699/0645.Set%20Mismatch/README_EN.md)
 - [Count Primes](/solution/0200-0299/0204.Count%20Primes/README_EN.md)
 - [Missing Number](/solution/0200-0299/0268.Missing%20Number/README_EN.md)

solution/0100-0199/0137.Single Number II/README.md

@@ -35,7 +35,19 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def singleNumber(self, nums: List[int]) -> int:
+        bits = [0] * 32
+        for num in nums:
+            for i in range(32):
+                bits[i] += (num & 1)
+                num >>= 1
+        res = 0
+        for i in range(32):
+            if bits[i] % 3 != 0:
+                res += (1 << i)
+        # 如果为负数，先将 0-32 位取反（即 res ^ 0xffffffff ），再将所有位取反（即 ~ ）
+        return res if bits[31] % 3 == 0 else ~(res ^ 0xffffffff)
 ```
 
 ### **Java**

solution/0100-0199/0137.Single Number II/README_EN.md

@@ -35,7 +35,18 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def singleNumber(self, nums: List[int]) -> int:
+        bits = [0] * 32
+        for num in nums:
+            for i in range(32):
+                bits[i] += (num & 1)
+                num >>= 1
+        res = 0
+        for i in range(32):
+            if bits[i] % 3 != 0:
+                res += (1 << i)
+        return res if bits[31] % 3 == 0 else ~(res ^ 0xffffffff)
 ```
 
 ### **Java**

solution/0100-0199/0137.Single Number II/Solution.py

@@ -1,9 +1,12 @@
 class Solution:
-    def singleNumber(self, nums):
-        """
-        :type nums: List[int]
-        :rtype: int
-        """
-        temp=set(nums)
-        res=(sum(temp)*3-sum(nums))//2
-        return res
+    def singleNumber(self, nums: List[int]) -> int:
+        bits = [0] * 32
+        for num in nums:
+            for i in range(32):
+                bits[i] += (num & 1)
+                num >>= 1
+        res = 0
+        for i in range(32):
+            if bits[i] % 3 != 0:
+                res += (1 << i)
+        return res if bits[31] % 3 == 0 else ~(res ^ 0xffffffff)

solution/0200-0299/0260.Single Number III/README.md

@@ -30,15 +30,42 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def singleNumber(self, nums: List[int]) -> List[int]:
+        xor = 0
+        for num in nums:
+            xor ^= num
+        # x & (-x) 是保留位中最右边 1 ，且将其余的 1 设位 0 的方法
+        diff = xor & (-xor)
+        a = b = 0
+        for num in nums:
+            if (num & diff) == 0:
+                a ^= num
+            else:
+                b ^= num
+        return [a, b]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int[] singleNumber(int[] nums) {
+        int xor = 0;
+        for (int num : nums) {
+            xor ^= num;
+        }
+        int diff = xor & (-xor);
+        int a = 0, b = 0;
+        for (int num : nums) {
+            if ((num & diff) == 0) a ^= num;
+            else b ^= num;
+        }
+        return new int[]{a, b};
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0260.Single Number III/README_EN.md

@@ -28,13 +28,40 @@
 ### **Python3**
 
 ```python
+class Solution:
+    def singleNumber(self, nums: List[int]) -> List[int]:
+        xor = 0
+        for num in nums:
+            xor ^= num
+        diff = xor & (-xor)
+        a = b = 0
+        for num in nums:
+            if (num & diff) == 0:
+                a ^= num
+            else:
+                b ^= num
+        return [a, b]
 
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int[] singleNumber(int[] nums) {
+        int xor = 0;
+        for (int num : nums) {
+            xor ^= num;
+        }
+        int diff = xor & (-xor);
+        int a = 0, b = 0;
+        for (int num : nums) {
+            if ((num & diff) == 0) a ^= num;
+            else b ^= num;
+        }
+        return new int[]{a, b};
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0260.Single Number III/Solution.java

@@ -1,18 +1,15 @@
 class Solution {
     public int[] singleNumber(int[] nums) {
-        int t = 0;
+        int xor = 0;
         for (int num : nums) {
-            t ^= num;
+            xor ^= num;
         }
-        int lastBit = t & -t;
-        int t1 = 0, t2 = 0;
+        int diff = xor & (-xor);
+        int a = 0, b = 0;
         for (int num : nums) {
-            if ((num & lastBit) != 0) {
-                t1 ^= num;
-            } else {
-                t2 ^= num;
-            }
+            if ((num & diff) == 0) a ^= num;
+            else b ^= num;
         }
-        return new int[]{t1, t2};
+        return new int[]{a, b};
     }
-}
+}

solution/0200-0299/0260.Single Number III/Solution.py

@@ -0,0 +1,13 @@
+class Solution:
+    def singleNumber(self, nums: List[int]) -> List[int]:
+        xor = 0
+        for num in nums:
+            xor ^= num
+        diff = xor & (-xor)
+        a = b = 0
+        for num in nums:
+            if (num & diff) == 0:
+                a ^= num
+            else:
+                b ^= num
+        return [a, b]