找出数组中唯一的重复元素

1-N放在N+1个元素的数组中，只有唯一的一个元素值重复，其他均只出现一次。每个数组元素只能访问一次，设计一个算法，将它找出来，不用辅助存储空间。这个问题其实就是编程之美上面机器故障的题的翻版，解法就是利用异或操作求解代码不难写:

/*    * Copyright (c) 2011 alexingcool. All Rights Reserved.    */#include <iostream>#include <iterator>#include <algorithm>#include <cassert>using namespace std;const int size = 1001;int array[size];void generateArray(int repeatNum){assert(repeatNum > 0 && repeatNum <= 1000);int i;for(i = 1; i < size; i++)array[i - 1] = i;array[i - 1] = repeatNum;}int getRepeatNum(int (&array)[size]){int result = 0;int temp = 0;for(int i = 0; i < size - 1; i++)result ^= i + 1;for(int i = 0; i < size; i++)temp ^= array[i];return (result ^ temp);}void main(){int number;cout << "enter the repeat number:" << endl;cin >> number;generateArray(number);int result = getRepeatNum(array);cout << "the repeat number is: " << result << endl;} 

结果如下：

下面同样是由异或求出机器故障的解： 

/*    * Copyright (c) 2011 alexingcool. All Rights Reserved.    */#include <iostream>#include <iterator>#include <algorithm>#include <cassert>using namespace std;const int size = 9;int array[size] = {1, 2, 3, 3, 5, 4, 4, 1, 5};int getResult(int (&array)[size]){int result = 0;for(int i = 0; i < size; i++)result ^= array[i];return result;}void main(){int result = getResult(array);cout << "the obstacle mechine: " << result << endl;}

结果：