找到重复的数

/*  http://weibo.com/1915548291/yDjra5ajO#1348465113389  一个大小为n的数组，里面的数都属于范围[0, n-1]，有不确定的重复元素，  找到至少一个重复元素，要求O(1)空间和O(n)时间。 */#include <stdlib.h>#include <assert.h>#include <time.h>#define NELEMS(a) (sizeof(a)/sizeof(a[0]))void swap(int *x, int *y){  int tmp = *x; *x = *y; *y = tmp;}/*  Term: "right place": a[i] = i, "wrong place": a[i] != i  the while loop executes at most n times, because each execution  puts an element from the "wrong place" into the "right place",  and there are at most n elements at "wrong place".  Side effect: parameter a is changed*/int find_duplicate(int a[], int n){  int i;  for(i = 0; i < n; i++) {    /* a[0]..a[i-1] are in right place */    while(a[i] != i) {      if (a[i] == a[a[i]])return a[i];      elseswap(&a[i],&a[a[i]]);    }  }  /* a[0]..a[n-1] are in right place     no duplicate one */  return -1;}void shuffle(int a[],int n){  while(--n > 0) {    int k = rand() % (n+1);    swap(&a[k],&a[n]);  }}int main(){  /* simple test */  int i;  srand(clock());  for(i = 0; i < 10000000; i++) {    int a[] = {0,1,2,3,4,5,6,7,8,8};    shuffle(a,NELEMS(a));    assert(8 == find_duplicate(a,NELEMS(a)));  }  return 0;}