HDU1157 POJ2388 Who's in the Middle

问题链接：HDU1157 POJ2388 Who's in the Middle。基础级练习题，用C语言编写程序。

问题简述：输入n，然后输入n个整数，求这n个整数中大小位于中间的数。

问题分析：使用分治法，用求第k大数算法实现。

程序说明：另外一种解法更加简洁，参见以下链接。

参考链接：POJ2388 HDU1157 Who's in the Middle。

AC的C语言程序如下：

/* HDU1157 POJ2388 Who's in the Middle */

#include <stdio.h>

#define MAXN 10000

int data[MAXN+1];

int split(int a[], int low, int high)
{
  int part_element = a[low];

  for (;;) {
    while (low < high && part_element <= a[high])
      high--;
    if (low >= high) break;
    a[low++] = a[high];

    while (low < high && a[low] <= part_element)
      low++;
    if (low >= high) break;
    a[high--] = a[low];
  }

  a[high] = part_element;
  return high;
}

// 非递归选择问题算法程序
int selectmink(int a[], int low, int high, int k)
{
  int middle;

  for(;;) {
      middle = split(a, low, high);
      if(middle == k)
          return a[k];
      else if(middle < k)
          low = middle+1;
      else /* if(middle > k) */
          high = middle-1;
  }
}

int main(void)
{
    int n, ans, i;

    while(scanf("%d", &n) != EOF) {
        for(i=0; i<n; i++)
            scanf("%d", &data[i]);

        ans = selectmink(data, 0, n - 1, n / 2);

        printf("%d
", ans);
    }

    return 0;
}