Single Number

目录

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?

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//第一题，无脑超时
public class Solution {
    public int singleNumber(int[] A) {
        List<Integer> resultList = new ArrayList<Integer>();
        List list = new ArrayList();
        for(int i=0;i<A.length;i++){
            Integer aiInteger = new Integer(A[i]);
            if(list.contains(aiInteger)){
                resultList.remove(aiInteger);
            }else{
                list.add(aiInteger);
                resultList.add(aiInteger);
            }
        }
        return resultList.get(0);
    }
}
//查看别人算法后：按位异或，所有重复项抵消
public class Solution {
    public int singleNumber(int[] A) {
        int sum = 0;
        for(int i = 0;i<A.length;i++){
            sum^=A[i];
        }
        return sum;
    }
}
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