在整数数组中寻找第K大的数

题目描述:在一个整数数组中,根据快排的思路,找出数组中第K大的数

思路:

利用快排思想,例如在49个元素中找第24大的元素,首先,进行一次快排,(将大的放到前半段,晓得放到后半段),假设得到的中轴为p判断p-low+1 == k,如果成立,直接输出a[p],(因为前半段有k-1个大于a[p]的元素,故a[p]为第k大的元素)如果p-low+1 > k,则第k大的元素在前半段,此时更新high = p-1,继续步骤2如果p-low+ < k,则第k大的元素在后半段,此时更新low = p+1,且k= k-(p=low+1),继续步骤2[时间复杂度为O(n)]

代码展示:

package com.bittech.Test; /** * package:com.bittech.Test * Description:TODO * @date:2019/5/26 * @Author:weiwei **/ public class Test0526 { public int findKth(int[] a, int n, int k) { return findKth(a, 0, n - 1, k); } public int findKth(int[] a, int low, int high, int k) { int part = partation(a, low, high); if (k == part - low + 1) { return a[part]; } else if (k > part - low + 1) { return findKth(a, part + 1, high, k - part + low - 1); } else { return findKth(a, low, part - 1, k); } } public int partation(int[] a, int low, int high) { int key = a[low]; while (low < high) { while (low < high && a[high] <= key) { high--; a[low] = a[high]; } while (low < high && a[low] >= key) { low++; a[high] = a[low]; } a[low] = key; return low; } return low; } }