[LeetCode]Single Number

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?

1. 不考虑多余内存空间，用HashMap解决很常规，时间复杂度为O(n)。

public class Solution {
public int singleNumber(int[] A) {
HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
int key,val;
val=0;
for(int i=0;i<A.length;i++){
if(map.containsKey(A[i])){
map.remove(A[i]);
}
else{
map.put(A[i],A[i]);
}
}
Iterator iter=map.keySet().iterator();
while(iter.hasNext()){
val=map.get(iter.next());
}
return val;
}
}

2. note里又说do it without using extra memory，这就难办了。想了很久没找到诀窍，最后求助万能互联网。

异或运算：^

4000

输入	输入	结果
1	1	0
1	0	1
0	1	1
0	0	0

代码变得十分简洁，太棒了！

public class Solution {
public int singleNumber(int[] A) {
for(int i=1;i<A.length;i++){
A[0]=A[0]^A[i];
}
return A[0];
}
}