Add explaination for "Single Number I/II"

Author: haoel

src/singleNumber/singleNumber.II.cpp

@@ -14,16 +14,71 @@
 
 class Solution {
 public:
+    Solution(){
+        srand(time(0));
+    }
+    
+    //random invoker
     int singleNumber(int A[], int n) {
+        if (rand()%2){
+            return singleNumber_1(A, n);
+        }
+        return singleNumber_2(A, n);
+    }
+
+    /*
+     *  This solution is clear & straightforward implementation.
+     *  
+     *  We use an array of 32 length(e.g. count[32]) to count the the bits for all of numbers. 
+     *
+     *  Because the same number appear 3 times, which means the sum of i-th bits for all numbers should be 3 times.
+     *
+     *  In other word, the sum of i-th bits mod 3, it must be 0 or 1. 1 means that is the single number bit.
+     *
+     * This solution can be easy to extend to "every element appears k times except for one."
+     *
+     */
+    int singleNumber_1(int A[], int n) {
+        int count[32] = {0};
+        int result = 0;
+        for (int i = 0; i < 32; i++) {
+            for (int j = 0; j < n; j++) {
+                if ((A[j] >> i) & 1) {
+                    count[i]++;
+                }
+            }
+            result |= ((count[i] % 3) << i);
+        }
+        return result;
+    }
+
+
+    /*
+     *   The following solution is popular solution on Internet, but it looks it's not easy to understand.
+     *
+     *   Actually, it just optimizes the above soultion.
+     *
+     *   Let's see how it improve the above.
+     *
+     *   We use three bitmask, 
+     *    1) `ones` represent the i-th bit had apear once.
+     *    2) `twos` represent the i-th bit had apear twice.
+     *    3) `threes` represent the i-th bit had apear three times.
+     *
+     *    When the i-th bit had appeared for the third time, clear the i-th bit of both `ones` and `twos` to 0.
+     *    The final answer will be the value of `ones`
+     *
+     */
+    int singleNumber_2(int A[], int n) {
         int ones = 0, twos = 0, threes = 0;
         for (int i = 0; i < n; i++) {
             twos |= ones & A[i];
-            ones ^= A[i];// 异或3次 和 异或 1次的结果是一样的
-            //对于ones 和 twos 把出现了3次的位置设置为0 （取反之后1的位置为0）
+            ones ^= A[i];
             threes = ones & twos;
             ones &= ~threes;
             twos &= ~threes;
         }
         return ones;
     }
+
 };

src/singleNumber/singleNumber.cpp

@@ -13,7 +13,9 @@
 **********************************************************************************/
 
 #include <stdio.h>
-
+// This is classical interview question
+// As we know, the same number XOR together will be 0,
+// So, XOR all of numbers, the result is the number which only appears once. 
 int singleNumber(int A[], int n) {
     int s = 0;
     for(int i=0; i<n; i++){