区间 top k

按编程之美上解法二的所述，类似快速排序的划分方法，N个数存储在数组S中，再从数组中随机选取一个数X，把数组划分为Sa和Sb俩部分，Sa<=X<=Sb，如果要查找的k个元素小于Sa的元素个数，则返回Sa中较小的k个元素，否则返回Sa中所有的元素+Sb中较小的k-|Sa|个元素。不断递归下去，把问题分解成更小的问题，平均时间复杂度为O（N）

// Copyright 2012 Jike Inc. All Rights Reserved.
// Author: jichenghui@jike.com(Chenghui Ji)
// Date  : 2012-08-24 16:35:51
// Brief : 
#include <iostream>
using namespace std;

int partition(int* data, int left, int right) {
  if (left > right) {
    return -1;
  } else {
    int polit = data[left];
    while(left < right) {
      while(right > left && data[right] >= polit) 
        right--;
      data[left] = data[right];
      while (left < right && data[left] <= polit)
        left++;
      data[right] = data[left];
    }
    data[left] = polit;
  }
  return left;
}

void topk_qsort(int* data, int left, int right, int k) {
  if (left >= right || k > right - left + 1) {
    return;
  }
  int polit =  partition(data, left, right);
  if (polit < left || polit >= right) return;
  int left_len = polit - left + 1;
  if (left_len == k) return;
  else if (left_len > k) {
    topk_qsort(data, left, polit-1, k);
  } else {
    topk_qsort(data, polit + 1, right, k - left_len);
  }
}
int main() {
  int data[] = {13,2,3,5,67,8,9,10,17,6};
  int len = sizeof(data)/sizeof(data[0]);
  cout << "orignal data:" << endl;
  for (int i = 0; i < len; ++i) {
    cout << data[i] << " ";
  }
  cout << endl;
  int k = 3;
  cout << "top" << k << "data:" << endl;
  topk_qsort(data, 0, len-1, k);
  for (int i = 0; i < len; ++i) {
    cout << data[i] << " ";
  }
  cout << endl;
  return 0;
}