标签：style blog http io color ar sp java for
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 import java.util.Hashtable; 2 public class Solution { 3 public int singleNumber(int[] A) { 4 Hashtable<Integer, Integer> hashtable = new Hashtable<Integer,Integer>(); 5 int result = 0; 6 7 for(int i = 0; i < A.length;i++){ 8 Integer element = hashtable.get(A[i]); 9 if(null == element){ 10 hashtable.put(A[i], 1);//第一次放1 11 } 12 else if(1 == element){ 13 hashtable.put(A[i], 2);//第二次放2 14 } 15 }//for 16 for(int i = 0; i < A.length; i++){ 17 if(1 == hashtable.get(A[i])){ 18 result = A[i]; 19 break; 20 } 21 }//for 22 return result; 23 } 24 }
ps
因为题目有要求，用O（1）空间，O（n）时间复杂度，用了hash表后，空间复杂度变为了O(n)
这里可以用异或来处理
异或具有交换性，a ^ b = b ^ a
0 ^ a = a
将所有的A[i]异或起来，A[0] ^ A[1]....这样根据交换性，最后只剩下单个的个数
只扫描了一遍，时间复杂度为O(N),只用了一个存储单元空间复杂度为O(1)
参考：http://www.cnblogs.com/changchengxiao/p/3413294.html
1 public class Solution { 2 public int singleNumber(int[] A) { 3 int result = 0; 4 for(int i = 0; i < A.length; i++){ 5 result ^= A[i]; 6 } 7 8 return result; 9 } 10 }
标签：style blog http io color ar sp java for
原文地址：http://www.cnblogs.com/luckygxf/p/4104371.html
