找到n個數中的最大的k個數

此題和算法導論9-1題很像，但是前k個大數不要求有序，所以可以省點事。利用快速排序的思想和修改算法導論9.2節線性時間選擇元素的RANDOMIZED-SELECT(),算法期望時間複雜度 O(n).

  public static int selectIndex(int[] array, int begin, int end, int kMax) {
    if( begin == end) {
      return begin;
    }
    int pivotIndex = partition(array, begin, end);
    int maxNb = pivotIndex - begin + 1;
    if(kMax == maxNb) {
      return pivotIndex;
    }
    else if( maxNb < kMax) {
      return selectIndex(array, pivotIndex+1, end, kMax-maxNb);
    }
    else {
      return selectIndex(array, begin, pivotIndex-1, kMax);
    }
  }

  public static void main(String[] args) {
    int[] array =  {3, 2, 9, 0, 7, 5, 4, 8, 6, 1 };
    int[] array1 = { 2, 8, 7, 1, 3, 5, 6, 4, 9};
    int[] array2 = { 13, 19, 9, 5, 12, 8, 7, 4, -1, 2, 6, 11 };

    selectMaxNumber(array, 8);
    selectMaxNumber(array1, 7);
    selectMaxNumber(array2, 10);
  }

  static int partition(int[] array, int start, int end) {
    int pivotValue = array[end];
    int i = start - 1;
    for (int q = start; q < end; q++) {
      if (array[q] > pivotValue) {
        i = i + 1;
        exchange(array, i, q);
      }
    }
    exchange(array, i + 1, end);
    return i + 1;
  }

  private static void exchange(int[] array, int i, int q) {
    int temp = array[i];
    array[i] = array[q];
    array[q] = temp;
  }

  public static void selectMaxNumber(int[] array, int kMax) {
    int pivotIndex = selectIndex(array, 0,  array.length-1, kMax);
    System.out.println("The "+ kMax + " numbers are:");
    for(int i=0; i<= pivotIndex; i++) {
      System.out.print(array[i] + " ");
    }
    System.out.println();
  }