LeetCode 137：Single Number II

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Single Number II 比Single Number要复杂的多，很难直观的找到算法。

考虑每个元素的为一个32位的二进制数，这样每一位上出现要么为1 ，要么为0。对数组，统计每一位上1 出现的次数count，必定是3N或者3N+1 次。让count对3取模，能够获得到那个只出现1次的元素该位是0还是1。

代码如下：

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int length = nums.size();
        int result = 0;
        for(int i = 0; i<32; i++){
            int count = 0; 
            int mask = 1<< i;
            for(int j=0; j<length; j++){
                if(nums[j] & mask)
                    count++;
            }
          if(count %3)
                result |= mask;
        }
        return result;
    }
};

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int length = nums.size();
        int result = 0;
        for(int i = 0; i<32; i++){
            int count = 0;
            int mask = 1<< i;
            for(int j=0; j<length; j++){
                if(nums[j] & mask)
                    count++;
            }
           if(count %2)
                result |= mask;
        }
        return result;
    }
};

对于Single Number II，网上还有一种与、异或等位操作的解法，尚未完全理解，先记录下

int singleNumber(int A[], int n)
{
	int one = 0, two = 0;
	for (int i = 0; i < n; i++)
	{
		two |= A[i] & one;
		one ^= A[i];
		int three = one & two;
		one &= ~three;
		two &= ~three;
	}
	return one;
}

LeetCode 137：Single Number II

原文：http://blog.csdn.net/sunao2002002/article/details/46318025