减治算法之寻找第K小元素问题

一、问题描写叙述

给定一个整数数列，寻找其按递增排序后的第k个位置上的元素。

二、问题分析

借助类似快排思想实现pation函数。再利用递归思想寻找k位置。

三、算法代码

public static int selectMinK(int [] arr, int low, int high, int k){
		int index = pation(arr, low, high);
		if(index == k){
			return arr[index];
		}
		if(index < k){
			return selectMinK(arr, index + 1, high, k);
		}else{
			return selectMinK(arr, low, index - 1, k);
		}
	}

	public static int pation(int [] arr, int low, int high){
		while(low < high){
			while(low < high && arr[low] <= arr[high]){//从后往前。把小的元素往前调换
				high--;
			}
			if(low < high){
				int tmp = arr[low];
				arr[low] = arr[high];
				arr[high] = tmp;
				low++;
			}
			while(low < high && arr[low] <= arr[high]){//从前往后。把大的元素往后调换
				low++;
			}
			if(low < high){
				int tmp = arr[low];
				arr[low] = arr[high];
				arr[high] = tmp;
				high--;
			}
		}
		return low;//返回low。high相遇位置
	}

四、完整測试代码

public class Solution {

	public static void main(String [] args){
		int [] randArr = new int[]{5,2,8,6,3,6,9,7};
		int result = selectMinK(randArr, 0, randArr.length - 1, 4);
		System.out.print(result);
	}
	public static int selectMinK(int [] arr, int low, int high, int k){
		int index = pation(arr, low, high);
		if(index == k){ //若返回的下标为k，则找到目标元素
			return arr[index];
		}
		if(index < k){
			return selectMinK(arr, index + 1, high, k);
		}else{
			return selectMinK(arr, low, index - 1, k);
		}
	}

	public static int pation(int [] arr, int low, int high){
		while(low < high){
			while(low < high && arr[low] <= arr[high]){
				high--;
			}
			if(low < high){
				int tmp = arr[low];
				arr[low] = arr[high];
				arr[high] = tmp;
				low++;
			}
			while(low < high && arr[low] <= arr[high]){
				low++;
			}
			if(low < high){
				int tmp = arr[low];
				arr[low] = arr[high];
				arr[high] = tmp;
				high--;
			}
		}
		return low;
	}
}

五、执行结果

第4小元素为：6