【Leetcode】Sort List (Sorting)

此题要求用归并排序排两个链表，基本思想还是分为分割和合并

合并的代码在Merge Two Sorted List里已经讲得很清楚了。所以这里直接给出代码。

	public ListNode merge(ListNode l1, ListNode l2) {
		ListNode helper = new ListNode(0);
		ListNode runner = helper;
		while (l1 != null && l2 != null) {
			if (l1.val < l2.val) {
				runner.next = l1;
				l1 = l1.next;
				runner = runner.next;
			} else {
				runner.next = l2;
				l2 = l2.next;
				runner = runner.next;
			}
			if (l1 != null)
				runner.next = l1;
			if (l2 != null)
				runner.next = l2;
		}
		return helper.next;
	}

分割的思想很重要，在数组里面是通过分割知道只剩下一个元素的时候结束，链表也一样。

数组中判断只有一个元素的方法是if(left==right)

链表中判断只有一个元素的方法是if(head.next==null)

数组中的分割的方法是

		int middle = (left + right) / 2;
		mergeSort(array, left, middle);
		mergeSort(array, middle + 1, right);
		merge(array, left, middle, right);

链表中怎么找middle呢？这个时候我们想起了前面所提到的Walker_Runner技术来找中点。

		ListNode walker = head;
		ListNode runner = head;
		while (runner.next != null && runner.next.next != null) {
			walker = walker.next;
			runner = runner.next.next;
		}
		ListNode head2 = walker.next;
		walker.next = null;
		head = mergeSort(head);
		head2 = mergeSort(head2);
		return merge(head, head2);

下面给出完整的代码

	public ListNode sortList(ListNode head) {
		if (head == null)
			return head;
		return mergeSort(head);
	}

	public ListNode mergeSort(ListNode head) {
		if (head.next == null)
			return head;
		ListNode walker = head;
		ListNode runner = head;
		while (runner.next != null && runner.next.next != null) {
			walker = walker.next;
			runner = runner.next.next;
		}
		ListNode head2 = walker.next;
		walker.next = null;
		head = mergeSort(head);
		head2 = mergeSort(head2);
		return merge(head, head2);
	}

	public ListNode merge(ListNode l1, ListNode l2) {
		ListNode helper = new ListNode(0);
		ListNode runner = helper;
		while (l1 != null && l2 != null) {
			if (l1.val < l2.val) {
				runner.next = l1;
				l1 = l1.next;
				runner = runner.next;
			} else {
				runner.next = l2;
				l2 = l2.next;
				runner = runner.next;
			}
			if (l1 != null)
				runner.next = l1;
			if (l2 != null)
				runner.next = l2;
		}
		return helper.next;
	}