寻找最小的整数

题目:给定一个无序的整数数组，怎么找到第一个大于0，并且不在此数组的整数。比如[1,2,0]返回3，[3,4,-1,1]返回2，[1, 5, 3, 4, 2]返回6，[100, 3, 2, 1, 6,8, 5]返回4。要求使用O(1)空间和O(n)时间

解答:不停的让a[i] =i  归位。

#include <stdio.h>
void Swap(int &a, int &b)
{
  int c = a;
  a = b;
  b = c;
}
int FindFirstNumberNotExistenceInArray(int a[], int n)
{
  int i;
  for (i = 0; i < n; i++)
    while (a[i] > 0 && a[i] <= n && a[i] != i + 1 && a[i] != a[a[i] - 1])
        Swap(a[i], a[a[i] - 1]);
  for (i = 0; i < n; i++)
    if (a[i] != i + 1)
      break;
  return i + 1;
}
void PrintfArray(int a[], int n)
{
  for (int i = 0; i < n; i++)
    printf("%d ", a[i]);
  putchar(‘\n‘);
}
int main()
{
 const int MAXN = 5;
  int a[MAXN] = {2, -100, 4, 1, 70};
  PrintfArray(a, MAXN);
  printf("该数组缺失的数字为%d\n", FindFirstNumberNotExistenceInArray(a, MAXN));
  return 0;
}

参考：http://blog.csdn.net/morewindows/article/details/12683723