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?
我的解题思路：不能使用额外空间，时间要线性，粗暴的方法就是两个循环
public class Solution {
public int singleNumber(int[] A) {
for (int i=0;i<A.length;i++){
Boolean twice = false;
for (int j= 0;j<A.length;j++){
if (i == j)
continue;
if (A[i] == A[j]){
twice = true;
break;
}
}
if (!twice){
return A[i];
}
}
return 0;
}
}
大神代码 时间150ms 空间12952k 异或的思路就是好啊，两个相同的数异或为0
public class Solution {
public int singleNumber(int[] A) {
int num = 0;
for(int i=0;i<A.length;i++){
num^=A[i];
}
return num;
}
}
原文：http://www.cnblogs.com/Melinni/p/7436466.html