找到重复的数
/*http://weibo.com/1915548291/yDjra5ajO#1348465113389一個大小為n的數組，里面的數都屬于范圍[0, n-1]，有不確定的重復元素，找到至少一個重復元素，要求O(1)空間和O(n)時間。*/#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] != ithe while loop executes at most n times, because each executionputs 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 placeno 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;
}
總結
