分治法 - Merge k Sorted Lists

一，Merge k Sorted Lists

题目描述

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

我的代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *merge(ListNode *list1, ListNode *list2){
        ListNode *p1, *p2, *head, *pre;
        p1 = list1;
        p2 = list2;
        head = new ListNode(-1);
        head->next = NULL;
        pre = head;
        while(p1 != NULL && p2 != NULL){
            if(p1->val < p2->val){
                pre->next = p1;
                pre = p1;
                p1 = p1->next;
            } else {
                pre->next = p2;
                pre = p2;
                p2 = p2->next;
            }
        }
        pre->next = NULL;
        if(p1 != NULL){
            pre->next = p1;
        }
        if(p2 != NULL){
            pre->next = p2;
        }
        return head->next;
    }

    /*
        函数的返回值是一个指针，指向一个单链表，这个单链表是由下标从l到r的单链表融合而成的
    */
    ListNode *getList(int l, int r, vector<ListNode*> &lists)
    {
        if(l == r){
            return lists[l];
        }
        if(l < r){
            int mid = (l + r) / 2;
            ListNode *left = getList(l, mid, lists);
            ListNode *right = getList(mid + 1, r, lists);
            return merge(left, right);
        }
        return NULL;
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        return getList(0, lists.size() - 1, lists);
    }
};