Java: 实现顺序表和单链表的快速排序

快速排序

• 快速排序原理

　　快速排序(Quick Sort)的基本思想是，通过一趟排序将待排记录分割成独立的两部分，其中一部分记录的关键字均比另一部分记录的关键字小，则可对这两部分记录继续进行排序，以达到整个序列有序的目的。

• 主程序

package Sort;

public class QuickSort {
	
	private void Qsort(int[] list, int low, int high){
		int privot;
		if(low < high) {
			print(list);
			/*算出枢轴值privot*/
			privot = Partition(list, low, high);
			/*对低子表递归排序*/
			Qsort(list, low, privot-1);
			/*对高子表递归排序*/
			Qsort(list, privot+1, high);			
		}
	}
	
	private int Partition(int[] list, int low, int high){
		/*用表的第一个记录做枢轴记录*/
		int privotkey = list[low];
		while (low < high){
			while (low < high && list[high] > privotkey)
				high--;			
			/*将比枢轴记录小的记录交换到低端*/
			swap(list, low, high);
			while (low < high && list[low] < privotkey)
				low++;
			/*将比枢轴记录大的记录交换到高端*/
			swap(list, low, high);			
		}
		/*返回枢轴所在的位置*/
		return low;
	}
	
	private void swap(int [] list,int j, int i){
		int temp;
		temp = list[i];
		list[i] = list[j];
		list[j] = temp;
	}	
	
	private void print(int[] data) {
		System.out.println("当前list数组的值：");
		for (int i = 0; i < data.length; i++) {
			System.out.print(data[i] + "\t");
		}
		System.out.println();
	}
	
	public static void main(String[] args) {
		int[] list = {50,40,80,30,60,70,10,90,20};
		QuickSort qs = new QuickSort();
		qs.Qsort(list, 0, list.length-1);
		qs.print(list);
	}
}

　　

• Qsort()分析

	private void Qsort(int[] list, int low, int high){
		int privot;
		if(low < high) {
			print(list);
			/*算出枢轴值privot*/
			privot = Partition(list, low, high);
			/*对低子表递归排序*/
			Qsort(list, low, privot-1);
			/*对高子表递归排序*/
			Qsort(list, privot+1, high);			
		}
	}

　　

• Partition()分析

	private int Partition(int[] list, int low, int high){
		/*用表的第一个记录做枢轴记录*/
		int privotkey = list[low];
		while (low < high){
			while (low < high && list[high] > privotkey)
				high--;			
			/*将比枢轴记录小的记录交换到低端*/
			swap(list, low, high);
			while (low < high && list[low] < privotkey)
				low++;
			/*将比枢轴记录大的记录交换到高端*/
			swap(list, low, high);			
		}
		/*返回枢轴所在的位置*/
		return low;
	}

　　

• 输出结果

当前list数组的值：
50	40	80	30	60	70	10	90	20	
当前list数组的值：
20	40	10	30	50	70	60	90	80	
当前list数组的值：
10	20	40	30	50	70	60	90	80	
当前list数组的值：
10	20	30	40	50	70	60	90	80	
当前list数组的值：
10	20	30	40	50	60	70	90	80	
当前list数组的值：
10	20	30	40	50	60	70	80	90	

　　

• 时间复杂度分析

　　最优情况下，快速排序算法的时间复杂度为O(nlogn)，空间复杂度也为O(nlogn)

•  单链表快速排序实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null) return head;
        
        ListNode left = new ListNode(0), leftHead = left;
        ListNode right = new ListNode(0), rightHead = right;
        ListNode mid = new ListNode(0), midHead = mid;
        int val = head.val;
        
        while(head != null) {
            if(head.val < val) {
                left.next = head;
                left = head;
            } else if(head.val > val) {
                right.next = head;
                right = head;
            } else {
                mid.next = head;
                mid = head;
            }
            head = head.next;
        }
        
        left.next = null;
        right.next = null;
        mid.next = null;
        return merge(sortList(leftHead.next), midHead.next, sortList(rightHead.next));
    }
    
    public ListNode merge(ListNode left, ListNode mid, ListNode right) {
        ListNode leftTail = getTail(left);
        ListNode midTail = getTail(mid);
        midTail.next = right;
        if(leftTail != null) {
            leftTail.next = mid;
            return left;
        } else {
            return mid;
        }
    }
    
    public ListNode getTail(ListNode head) {
        if(head == null) return head;
        while(head.next != null) {
            head = head.next;
        }
        return head;
    }
}