[leetcode]Sort List

merge sort，记了长度。如果断开链表时把第一个链表的最后一个节点的next设为NULL，就不用在函数参数里传长度了。一直错误是因为在生成的链表最后的next没有设成NULL。用了dummy node很有用。

可以借鉴http://blog.csdn.net/sunbaigui/article/details/16843419 最后直接把剩余的链表接过来。

class Solution {

public:

    ListNode *sortList(ListNode *head) {

        int size = 0;

        ListNode *node = head;

        while (node != NULL) {

            node = node->next;

            size++;

        }

        return mergeSort(head, size);

    }

    

    ListNode* mergeSort(ListNode *node, int size) {

        if (size == 0 || size == 1)

            return node;

        int half = size / 2;

        ListNode* head1 = node;

        ListNode* head2 = node;

        for (int i = 0; i < half; i++) {

            head2 = head2->next;

        }

        head1 = mergeSort(head1, half);

        head2 = mergeSort(head2, size - half);



        ListNode *dummy = new ListNode(0);

        ListNode *prev = dummy;

        ListNode *n1 = head1;

        ListNode *n2 = head2;

        int n1Step = 0;

        int n2Step = 0;

        while (n1Step != half && n2Step != size - half) {

            if (n1->val <= n2->val) {

                prev->next = n1;

                n1 = n1->next;

                n1Step++;

            }

            else {

                prev->next = n2;

                n2 = n2->next;

                n2Step++;

            }

            prev = prev->next;

        }

        while (n2Step != size - half) {

            prev->next = n2;

            prev = prev->next;

            n2 = n2->next;

            n2Step++;

        }

        while (n1Step != half) {

            prev->next = n1;

            prev = prev->next;

            n1 = n1->next;

            n1Step++;

        }

        prev->next = NULL;

        return dummy->next;

    }

};