Add solution 137

Author: yanglbme

README.md

@@ -61,6 +61,7 @@ Complete solutions to Leetcode problems, updated daily.
 | 102 | [Binary Tree Level Order Traversal](https://github.com/doocs/leetcode/tree/master/solution/102.Binary%20Tree%20Level%20Order%20Traversal) | `Tree`, `Breadth-first Search` |
 | 127 | [Word Ladder](https://github.com/doocs/leetcode/tree/master/solution/127.Word%20Ladder) | `Breadth-first Search` |
 | 130 | [Surrounded Regions](https://github.com/doocs/leetcode/tree/master/solution/130.Surrounded%20Regions) | `Depth-first Search`, `Breadth-first Search`, `Union Find` |
+| 137 | [Single Number II](https://github.com/doocs/leetcode/tree/master/solution/137.Single%20Number%20II) | `Bit Manipulation` |
 | 144 | [Binary Tree Preorder Traversal](https://github.com/doocs/leetcode/tree/master/solution/144.Binary%20Tree%20Preorder%20Traversal) | `Stack`, `Tree` |
 | 150 | [Evaluate Reverse Polish Notation](https://github.com/doocs/leetcode/tree/master/solution/150.Evaluate%20Reverse%20Polish%20Notation) | `Stack` |
 | 153 | [Find Minimum in Rotated Sorted Array](https://github.com/doocs/leetcode/tree/master/solution/153.Find%20Minimum%20in%20Rotated%20Sorted%20Array) | `Array`, `Binary Search` |

solution/137.Single Number II/README.md

@@ -0,0 +1,46 @@
+## 只出现一次的数字 II
+
+### 题目描述
+给定一个**非空**整数数组，除了某个元素只出现一次以外，其余每个元素均出现了三次。找出那个只出现了一次的元素。
+
+说明：
+
+你的算法应该具有线性时间复杂度。 你可以不使用额外空间来实现吗？
+
+**示例 1:**
+```
+输入: [2,2,3,2]
+输出: 3
+```
+
+**示例 2:**
+```
+输入: [0,1,0,1,0,1,99]
+输出: 99
+```
+
+### 解法
+遍历数组元素，对于每一个元素，获得二进制位（0/1），累加到 bits 数组中，这样下来，出现三次的元素，bits 数组上的值一定能被 3 整除；找出不能被 3 整除的位，计算出实际的十进制数即可。
+
+```java
+class Solution {
+    public int singleNumber(int[] nums) {
+        int[] bits = new int[32];
+        int n = nums.length;
+        for (int i = 0; i < n; ++i) {
+            for (int j = 0; j < 32; ++j) {
+                bits[j] += ((nums[i] >> j) & 1);
+            }
+        }
+        
+        int res = 0;
+        for (int i = 0; i < 32; ++i) {
+            if (bits[i] % 3 != 0) {
+                res += (1 << i);
+            }
+        }
+        return res;
+        
+    }
+}
+```

solution/137.Single Number II/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public int singleNumber(int[] nums) {
+        int[] bits = new int[32];
+        int n = nums.length;
+        for (int i = 0; i < n; ++i) {
+            for (int j = 0; j < 32; ++j) {
+                bits[j] += ((nums[i] >> j) & 1);
+            }
+        }
+        
+        int res = 0;
+        for (int i = 0; i < 32; ++i) {
+            if (bits[i] % 3 != 0) {
+                res += (1 << i);
+            }
+        }
+        return res;
+        
+    }
+}