leetcode.136. Single Number

Question

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

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

Solution

int compare( const void* a, const void* b)
{
     int int_a = * ( (int*) a );
     int int_b = * ( (int*) b );

     if ( int_a == int_b ) return 0;
     else if ( int_a < int_b ) return -1;
     else return 1;
}

int singleNumber(int* nums, int numsSize) {
    qsort( nums, numsSize, sizeof(int), compare );
    int i;
    for (i = 0; i < numsSize; i+=2) {
        if (nums[i]!=nums[i+1]) {
           return nums[i];
        }
    }
    return nums[numsSize-1];
}

Solution best of others(bit manipulation)

异或消除相同的位，剩下的就是single number

int singleNumber(int* nums, int numsSize) {
    int x = 0;
    while (numsSize--)
    {
        x ^= nums[numsSize];
    }
    return x;
}