leetcode single-number

https://oj.leetcode.com/problems/single-number/点击打开链接

Given an array of integers, every element appears twice except for one. Find that single one.

解题思想：利用异或运算的特点 A ^ A = 0    A ^ 0  = a

int singleNumber(int A[], int n)
    {
        int result = 0;
        for (int i = 0; i < n; i++)
        {
            result = result ^ A[i];
        }
        return result;
    }

https://oj.leetcode.com/problems/single-number-ii/ 点击打开链接

Problem description: Given an array of integers, every element appears three times except for one. Find that single one.

解题思路：该题目中处理1个元素外所有的元素都出现了3次，二进制异或的方法已经失效了。假设 int 是 32bit，对于 bit0 将所有数字的 bit0 相加得到 sum0。 sum0 % 3 就得到那个数字的 bit0的值，因为 (1 + 1 + 1) % 3 == 0, (0 + 0 + 0) % 3 == 0

int singleNumber(int A[], int n) 
    {
        //int nbits = sizeof(int) * 8;
        int result = 0;
        int count[32] = {0}; /* 此处采用了硬编码 */
        for (int i = 0; i < 32; i++)
        {
            for (int j = 0; j < n; j++)
            {
                count[i] += ((A[j] >> i) & 1);
            }
            result |= ((count[i] % 3) << i);
        }
        return result;
    }