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?
分析:位运算。
这个比Single Number要简单一点,只要将当前结果与遇到的数按位求异或(按位求异或,则该位出现两个1或者两个0得到的结果都是0,这样可以将出现两次的数清0,则剩下的必然是只出现了一次的那个数)。
public class Solution {
public int singleNumber(int[] A) {
int result = 0;
for(int i=0; i<A.length; i++){
result = result ^ A[i];
}
return result;
}
}