Single Number

Single Number I

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?

Single Number II

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

思路：

其实是一个小技巧。一个整数和它本身异或之后得到值是0。所以初始化一个值为0的变量，让数组中的所有数与之异或，然后就可以找到这个只出现一次的数。

题目可以扩展到寻找唯一的一个只出现奇数次的数。一样的方法。

对于第二个问题，因为只能用O(1)的空间，所以技巧是对每一个位的1的个数进行计数。这样唯一的只出现一次的数用到的位将导致计数不是3的倍数。最后检查所有计数不是3倍数的位，即可恢复原来的数字。

解答：

I.

   static int singleNumber(int A[], int n) {        if (n == 0) return 0;        int x = A[0];        for (int i = 1; i < n; i++) {        x^=A[i];        }        return x;    }

II.

static int singleNumber2(int A[], int n) {          int INTLEN = 4 * 8;          int[] bitCount = new int[INTLEN];          boolean f ;        for(int i = 0; i < n; ++i)              for(int j = 0; j < INTLEN; ++j)  {                 f = ((A[i] & (1 << j)) != 0);                   if(f) bitCount[j] += 1;   <span style="font-family: Arial;">//对A的每个元素对应二进制的每一位1的个数进行统计</span>            }        int single = 0;          for(int j = 0; j < INTLEN; ++j)              if ((bitCount[j] % 3) == 1)                  single += (1 << j);          return single;      }  

题目变形：

时间限制 1000 ms 内存限制 65536 KB

题目描述

Given an array with N integers where all elements appear three times except for one. Find out the one which appears only once.

输入格式

Several test cases are given, terminated by EOF.

Each test case consists of two lines. The first line gives the length of array N(1≤N≤105), and the other line describes the Nelements. All elements are ranged in [0,263−1].

输出格式

Output the answer for each test case, one per line.

解答：

<pre name="code" class="java">import java.math.BigInteger;import java.util.Scanner;public class Main {static BigInteger singleNumber(BigInteger A[], int n) {      int INTLEN = 64;      int[] bitCount = new int[INTLEN];      boolean f ;    for(int i = 0; i < n; ++i)      for(int j = 0; j < INTLEN; ++j)  {    f = (!(A[i].and(BigInteger.ONE.shiftLeft(j))).equals(BigInteger.ZERO));      if(f) bitCount[j] += 1;    }        BigInteger single = new BigInteger("0");      for(int j = 0; j < INTLEN; ++j)      if ((bitCount[j] % 3) == 1)      single = single.add(BigInteger.ONE.shiftLeft(j));      return single;      }  public static void main(String[] args) {Scanner sc = new Scanner(System.in);BigInteger[] bi ; int n ;while(sc.hasNext()){n = sc.nextInt();bi = new BigInteger[n];for(int i=0;i<n;i++){bi[i] = new BigInteger(sc.next());}System.out.println(singleNumber(bi, n).toString());}}}

参考网址：http://blog.csdn.net/magisu/article/details/13169283

http://code.bupt.edu.cn/problem/p/84/