leetcode 30天挑戰

Single Number (_XOR).c

//Example 1:
//Input: nums = [2,2,1]
//Output: 1

//Example 2:
//Input: nums = [4,1,2,1,2]
//Output: 4

//Example 3:
//Input: nums = [1]
//Output: 1

//way1
int singleNumber(int* nums, int numsSize){
    for(int i=0;i<numsSize;i++){
        int count=0;
        for(int j=0;j<numsSize;j++){
            if(nums[j]==nums[i]){
                count++;
            }
        }
        if(count==1) 
            return nums[i];
    }
    return 0;
}

//way2
int singleNumber(int* nums, int numsSize){
    int n=nums[0];
    for(int i=1;i<numsSize;i++){
        n^=nums[i];
      //n^nums[i]=nums[i];
      //nums:[2,2,1]
      //      2^2^1 = 0^1 = 1
      //   00000000 00000000 0000000 00000010
      // ^ 00000000 00000000 0000000 00000010
      //---------------------------------------
      //   00000000 00000000 0000000 00000000
        
      // 以上相同數會為0 A^A => 0 , A^0=>A
      //n= nums[0] ^ nums[1] ^ nums[2] ^ nums[3] = 2^2^1^2= 1 
      // bitwise XOR 每個bit分開計算
    }
    return n;
}