领扣排序链表之归并排序

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
        void merge(int a[], int left, int middle, int right)//进行排序
	{
		int *b = new int[right - left + 1];

		int i = left, j = middle + 1, k = 0;
		while (i <= middle && j <= right)//进行分组排序--左右两组之间的比较排序--建立在左右分组已经排好序的基础上
		{
			if (a[i]<a[j])//进行比较，将小的存入b[]中
				b[k++] = a[i++];
			else
				b[k++] = a[j++];
		}
		if (i>middle)//当右半部分仍有剩余时，将剩下的数存入b[]尾部
			while (j <= right)
				b[k++] = a[j++];
		else//当左半部分仍有剩余时，将剩下的数存入b[]尾部
			while (i <= middle)
				b[k++] = a[i++];
		j = 0;
		for (int i = left; i <= right; i++)//将b[]的值赋给a[]
		{
			a[i] = b[j];
			j++;
		}
	}
	void merge_sort(int left, int right, int a[])//分而治之，递归分组
	{
		if (left<right)
		{
			int middle = (left + right) / 2;
			merge_sort(left, middle, a);//递归分组左半部分
			merge_sort(middle + 1, right, a);//递归分组右半部分
			merge(a, left, middle, right);
		}
	}
	ListNode* sortList(ListNode* head) {
		if (head == NULL)//如果链表为空，直接返回
			return 0;
		int count = 0;
		ListNode* q = head;
		while (head != NULL)//求链表非空结点数
		{
			count++;
			head = head->next;
		}
		int *p = new int[count];
		head = q;
		 for (int i = 0; i<count; i++)//将链表每一个结点的值存入链表
		{
			p[i] = head->val;
			head = head->next;
		}
		merge_sort(0, count - 1, p);
		head = q;
		for (int i = 0; i<count; i++)//将排序好的中介数组存入链表
		{
			head->val = p[i];
			head = head->next;
		}
		return q;
	}
};

击败96.88%.