Single Number III(LintCode)

Single Number III

Given 2*n + 2 numbers, every numbers occurs twice except two, find them.

Example

Given [1,2,2,3,4,4,5,3] return 1 and 5

Challenge

O(n) time, O(1) extra space.

 

 

可耻的百度了。。点这里

还是好理解的。当然不一定要用最低位的1，任意找一个1即可。不过有lowbit = xor & -xor 这个公式当然是这个比较快。。

public class Solution {
    /**
     * @param A : An integer array
     * @return : Two integers
     */
    public List<Integer> singleNumberIII(int[] A) {
        List<Integer> list = new ArrayList<Integer>();
        
        int x = 0;
        for(int a : A) {
            x = x ^ a;
        }
        int lowbit = x & (-x);
        int a = 0;
        int b = 0;
        for(int y : A) {
            if((y & lowbit) == 0) {
                a = a ^ y;
            }else {
                b = b ^ y;
            }
        }
        
        
        list.add(a);
        list.add(b);
        return list;
    }
}