[LeetCode]Single Number
2014-07-01 19:27
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题目:给定一个数字集合,集合中除了数字X是唯一以外,其余数字均出现两次,在集合中找出X
算法:二进制,异或^操作符
原理:两个相同的数异或结果为0,因此在N个数字中,任意两个相同的数字异或结果为0,任何数A与0异或结果仍然为A
算法:二进制,异或^操作符
原理:两个相同的数异或结果为0,因此在N个数字中,任意两个相同的数字异或结果为0,任何数A与0异或结果仍然为A
public class SingleNumber { /* public static void main(String[] args) { // TODO Auto-generated method stub int[] A = new int[]{2,2,3,4,4,5,5,6,6,8,8}; Solution obj = new Solution(); int result = obj.singleNumber(A); System.out.println(result); } */ public static class Solution { public int singleNumber(int[] A) { int result = A[0]; for (int i=1; i<A.length; ++i) { result ^= A[i]; } return result; } } }
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