[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;
    }
};