leetCode解题报告之Sort List

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题目：

Sort a linked list in O(n log n) time using constant space complexity.

分析：

题目要求我们要用一个复杂度O(nlogn)的排序算法来排序一个链表，复杂度O(nlogn)的排序算法包括：快速排序，堆排序，希尔排序，二叉排序树排序，归并排序

情况：

考虑到题目的要求，我个人觉得用“归并排序”会比较好！

第一次提交没AC的Time Limit Exceeded代码：(没有考虑到两个递归之后的子链表做排序的话，关于合并的时间消耗)

package cn.xym.leetcode.sortlist;


public class Solution1 {
	
	public ListNode sortList(ListNode head) {
		if (head == null || head.next == null)
			return head;
		int len = getListSize(head);
		mergerSort(head, 0, len-1);
		
		return head;
	}
	public void mergerSort(ListNode head, int left, int right){
		int mid = (right+left) / 2;
		if (left == right)
			return;
		else{
			int first = left;
			int last = right;
			mergerSort(head, first, mid);
			mergerSort(head, mid+1, last);
			mergerListNode(head, first, mid, last);
		}
	}
   public void mergerListNode(ListNode head, int left,int mid,int right){
		int left1 = 0;
		int right1 = mid - left;
		int left2 = 0;
		int right2 = right - mid - 1;
		int k = 0;
		
		ListNode node1 = getListNode(head, left);
		ListNode node2 = getListNode(head, mid+1);
		int[] temp = new int[right - left + 1];
		
		while (left1 <= right1 && left2 <= right2){
			ListNode temp1 = getListNode(node1, left1);
			ListNode temp2 = getListNode(node2, left2);
			if (temp1.val < temp2.val){
				temp[k++] = temp1.val;
				left1++;
			}else{
				temp[k++] = temp2.val;
				left2++;
			}
		}
		while (left1 <= right1){
			temp[k++] = getListNode(node1, left1).val;
			left1++;
		}
		while (left2 <= right2){
			temp[k++] = getListNode(node2, left2).val;
			left2++;
		}
		for (int i=0; i<k; ++i){
			getListNode(head, i + left).val = temp[i];
		}
	}
	
	public ListNode getListNode(ListNode head, int index){
		ListNode temp = head;
		for (int i=0; i<index; ++i){
			temp = temp.next;
		}
		return temp;
	}
	public int getListSize(ListNode head){
		int len = 0;
		while (head != null){
			len++;
			head = head.next;
		}
		return len;
	}
	
	public static void main(String[] args) {
		ListNode head1 = new ListNode(10);
		ListNode head2 = new ListNode(40);
		ListNode head3 = new ListNode(6);
		ListNode head4 = new ListNode(22);
		ListNode head5 = new ListNode(1);
		ListNode head6 = new ListNode(30);
		head1.next = head2;
		head2.next = head3;
		head3.next = head4;
		head4.next = head5;
		head5.next = head6;
		head6.next = null;
		Solution1 solution = new Solution1();
		solution.sortList(head1);
		while (head1 != null){
			System.out.println(head1.val);
			head1 = head1.next;
		}
	}
}

下面是AC的代码：

package cn.xym.leetcode.sortlist;

class ListNode {
	int val;
	ListNode next;
	ListNode(int x) {
		val = x;
		next = null;
	}
}

public class Solution {

	public ListNode sortList(ListNode head) {
		if (head == null || head.next == null)
			return head;
		int len = getSize(head);
		ListNode list = mergerSort(head,len);
		return list;
	}

	public ListNode mergerSort(ListNode head, int len) {
		
		if (len == 1)
			return head;
		//找出当前head链表中间的ListNode
		int middle = len / 2;
		ListNode preMiddleNode = getListNode(head, middle-1);
		ListNode endNode = getListNode(head, len-1);
		ListNode middleNode = preMiddleNode.next;
		preMiddleNode.next = null;
		endNode.next = null;
		
		int leftLen = getSize(head);
		int rightLen = getSize(middleNode);
		
		ListNode leftList = mergerSort(head, leftLen);
		ListNode rightList = mergerSort(middleNode, rightLen);
		ListNode allList = mergerListNode(leftList, rightList);
		return allList;
	}
	/*返回一个局部排序好的表*/
	public ListNode mergerListNode(ListNode left, ListNode right) {
		ListNode temp = new ListNode(Integer.MIN_VALUE);
		ListNode preHead = new ListNode(Integer.MIN_VALUE);
		preHead = temp;
		
		//两个排序好的链表进行合并
		while (left != null && right != null){
			if (left.val < right.val){
				temp.next = left;
				left = left.next;
			}else{
				temp.next = right;
				right = right.next;
			}
			temp = temp.next;
		}
		if (left == null){
			temp.next = right;
		}
		if (right == null){
			temp.next = left;
		}
		
		return preHead.next;
	}

	public ListNode getListNode(ListNode head, int index) {
		ListNode temp = head;
		for (int i = 0; i < index; ++i) {
			temp = temp.next;
		}
		return temp;
	}

	public int getSize(ListNode head) {
		int len = 0;
		while (head != null) {
			len++;
			head = head.next;
		}
		return len;
	}

	public static void main(String[] args) {
		ListNode head1 = new ListNode(10);
		ListNode head2 = new ListNode(40);
		ListNode head3 = new ListNode(6);
		ListNode head4 = new ListNode(22);
		ListNode head5 = new ListNode(1);
		ListNode head6 = new ListNode(30);
		head1.next = head2;
		head2.next = head3;
		head3.next = head4;
		head4.next = head5;
		head5.next = head6;
		head6.next = null;
		Solution solution = new Solution();
		head1 = solution.sortList(head1);
		while (head1 != null) {
			System.out.println(head1.val);
			head1 = head1.next;
		}
	}
}

leetCode解题报告之Sort List,布布扣,bubuko.com

leetCode解题报告之Sort List

原文：http://blog.csdn.net/ljphhj/article/details/21317387