算法设计与分析基础14快速排序

Posted 2020-09-22 cutter_point

package cn.xf.algorithm.ch04;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

import org.junit.Test;

public class QuikSort {
	
	public int hoarePartition(List data, int left, int right) {
		//当要比对的数据相差为1的时候，表示就只有一条数据要进行比较了
		if(data == null || right - left <= 1 || left >= right) {
			return left;
		}
		
		int sour = (Integer) data.get(left); //以第一个元素起始作为分裂点
		int i = left;	//左边起始位置
		int j = right; //右边起始遍历位置
		
		do {
			//只要i在j的左边，就不断循环，知道i >= j
			do {
				++i;
			} while((Integer) data.get(i) < sour);
			//只要j在i右边就循环
			do {
				--j;
			} while((Integer) data.get(j) > sour);
			this.swap(data, i, j);
		} while(i < j);
		
		//交换完毕之后，吧分裂点后置到中间点
		this.swap(data, i, j);
		this.swap(data, left, j);
		//
		return j;
	}
	
	/**
	 * 交换数据
	 * @param data
	 * @param index1
	 * @param index2
	 */
	public void swap(List data, int index1, int index2) {
		int temp = (Integer) data.get(index1);
		data.set(index1, data.get(index2));
		data.set(index2, temp);
	}
	
	public void quikS(List data, int left, int right) {
		if(left < right) {
			int mid = this.hoarePartition(data, left, right);
			quikS(data, left, mid);
			quikS(data, mid + 1, right);
		}
	}
	
	@Test
	public void test2() {
		List data = Arrays.asList(5,3,1,9,8,2,4,7);
		quikS(data, 0, data.size());
		for(Object o : data) {
			System.out.print(o + "\\t");
		}
	}
	
	public static void test1(List data) {
		data = null;
	}
	
	@Test
	public void test() {
		List data = Arrays.asList("1", "2", "3", "4", "5", "6");
		QuikSort.test1(data);
		for(Object j : data) {
			System.out.println(j);
		}
	}
}

　　

显示结果：