Single Number(JAVA)

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?

 public class Solution {
     public int singleNumber(int[] A) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
      if(A.length == 0 || A == null){
          return 0;
      }
      for(int i=0;i<A.length;i++){
          if(map.containsKey(A[i])){
              int value = map.get(A[i]);
              map.put(A[i], value+1);
          }else {
              map.put(A[i], 1);
          }
      }
      for(int i=0;i<A.length;i++){
          int value = map.get(A[i]);
          if(value == 1){
              return A[i];
          }
      }
      return 0;
     }
 }

解法2

异或：相同为0 不同为1

0和a数异或都是a

a和a异或就是0

由于每个数有两个 所以总会相见一次 所以还是0

 public class singleNumber {

     /**
      * @param args
      */
     public static void main(String[] args) {
         int a[]={1,-1,3,3,4,1,5,5,-1,6};
         System.out.println(singleNumber(a));

     }

     // 异或
     public static int singleNumber(int[] A) {
         int res = 0;
         for (int i : A) {
             res ^= i;
         }
         return res;
     }

 }

屌爆