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
View Code
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。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 public class Solution { 2 /** 3 * @param A : An integer array 4 * @return : Two integers 5 */ 6 public List<Integer> singleNumberIII(int[] A) { 7 List<Integer> list = new ArrayList<Integer>(); 8 9 int x = 0; 10 for(int a : A) { 11 x = x ^ a; 12 } 13 int lowbit = x & (-x); 14 int a = 0; 15 int b = 0; 16 for(int y : A) { 17 if((y & lowbit) == 0) { 18 a = a ^ y; 19 }else { 20 b = b ^ y; 21 } 22 } 23 24 25 list.add(a); 26 list.add(b); 27 return list; 28 } 29 }