[LeetCode] Single Number

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?

href:  https://oj.leetcode.com/problems/single-number/

解题思路：两个相同的数异或为0，然后分为正负两种情况考虑，就得到线性的解法。

class Solution {
public:
    int singleNumber(int A[], int n) {
        int zres=0;
        int fres=0;
        for(int i=0;i<n;i++){
            if(A[i]>=0){
               zres ^=A[i]; 
            }else{
               fres ^=-A[i];
            }
            
        }
        if(0==zres&&0==fres){
            return 0;
        }else if(zres!=0){
            return zres;
        }else{
            return -fres;
        }
    }
};

public class Solution {
    public int singleNumber(int[] A) {
        int zres=0;
        int fres=0;
        for(int i=0;i<A.length;i++){
            if(A[i]>=0){
               zres ^=A[i]; 
            }else{
               fres ^=-A[i];
            }
            
        }
        if(0==zres&&0==fres){
            return 0;
        }else if(zres!=0){
            return zres;
        }else{
            return -fres;
        }
    }
}