LeetCode刷題筆記（鏈表）：merge-k-sorted-lists

---

• 轉載請註明作者和出處：http://blog.csdn.net/u011475210
• 代碼地址：https://github.com/WordZzzz/Note/tree/master/LeetCode
• 刷題平臺：https://www.nowcoder.com/ta/leetcode
• 題  庫：Leetcode經典編程題
• 編  者：WordZzzz

---

題目描述

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

解題思路

之前做過一道題是對兩個排序鏈表進行合併，這次是k個。所以，需要在原來的基礎長做一些封裝。

按照歸併排序的思想，我們通過二分法得到中點進行遞歸，自下而上進行兩兩歸併，直到只剩最後一個鏈表。

C++版代碼實現

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        if(lists.size() == 0)
            return NULL;
        return mergeList(lists, 0, lists.size()-1);
    }

    ListNode *mergeList(vector<ListNode *> &lists, int low, int high){
        if(high <= low)
            return lists[low];
        int mid = (low + high) / 2;
        ListNode *left = mergeList(lists, low, mid);
        ListNode *right = mergeList(lists, mid+1, high);
        return merge(left, right);
    }

    ListNode *merge(ListNode *left, ListNode *right) {
        if(left == NULL)
            return right;
        if(right == NULL)
            return left;

        ListNode *dummy = new ListNode(0);
        ListNode *cur = dummy;
        while(left != NULL && right != NULL){
            if(left->val < right->val){
                cur->next = left;
                left = left->next;
            }
            else{
                cur->next = right;
                right = right->next;
            }
            cur = cur->next;
        }
        cur->next = left ? left : right;
        return dummy->next;
    }
};

系列教程持續發佈中，歡迎訂閱、關注、收藏、評論、點贊哦～～(￣▽￣～)～

完的汪(∪｡∪)｡｡｡zzz