Java: 實現順序表和單鏈表的快速排序

快速排序

• 快速排序原理

　　快速排序(Quick Sort)的基本思想是，經過一趟排序將待排記錄分割成獨立的兩部分，其中一部分記錄的關鍵字均比另外一部分記錄的關鍵字小，則可對這兩部分記錄繼續進行排序，以達到整個序列有序的目的。

• 主程序

package Sort;

public class QuickSort {
	
	private void Qsort(int[] list, int low, int high){
		int privot;
		if(low < high) {
			print(list);
			/*算出樞軸值privot*/
			privot = Partition(list, low, high);
			/*對低子表遞歸排序*/
			Qsort(list, low, privot-1);
			/*對高子表遞歸排序*/
			Qsort(list, privot+1, high);			
		}
	}
	
	private int Partition(int[] list, int low, int high){
		/*用表的第一個記錄作樞軸記錄*/
		int privotkey = list[low];
		while (low < high){
			while (low < high && list[high] > privotkey)
				high--;			
			/*將比樞軸記錄小的記錄交換到低端*/
			swap(list, low, high);
			while (low < high && list[low] < privotkey)
				low++;
			/*將比樞軸記錄大的記錄交換到高端*/
			swap(list, low, high);			
		}
		/*返回樞軸所在的位置*/
		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("當前list數組的值：");
		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);
			/*算出樞軸值privot*/
			privot = Partition(list, low, high);
			/*對低子表遞歸排序*/
			Qsort(list, low, privot-1);
			/*對高子表遞歸排序*/
			Qsort(list, privot+1, high);			
		}
	}

　　

• Partition()分析

	private int Partition(int[] list, int low, int high){
		/*用表的第一個記錄作樞軸記錄*/
		int privotkey = list[low];
		while (low < high){
			while (low < high && list[high] > privotkey)
				high--;			
			/*將比樞軸記錄小的記錄交換到低端*/
			swap(list, low, high);
			while (low < high && list[low] < privotkey)
				low++;
			/*將比樞軸記錄大的記錄交換到高端*/
			swap(list, low, high);			
		}
		/*返回樞軸所在的位置*/
		return low;
	}

　　

• 輸出結果

當前list數組的值：
50	40	80	30	60	70	10	90	20	
當前list數組的值：
20	40	10	30	50	70	60	90	80	
當前list數組的值：
10	20	40	30	50	70	60	90	80	
當前list數組的值：
10	20	30	40	50	70	60	90	80	
當前list數組的值：
10	20	30	40	50	60	70	90	80	
當前list數組的值：
10	20	30	40	50	60	70	80	90	

　　

• 時間複雜度分析

　　最優狀況下，快速排序算法的時間複雜度爲O(nlogn)，空間複雜度也爲O(nlogn)

•  單鏈錶快速排序實現

/**
 * 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;
    }
}