LeetCode 23 合并K个排序链表

23. 合并K个排序链表

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

虽然说是一道困难题 但是真的很简单，

方法一：用vector保存数据 然后排序 然后生成链表就好啦

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector& lists) {
        ListNode *dummy = new ListNode(-1), *cur = dummy;
        vector v1({});
        for (auto node : lists) {
            ListNode *t = node;
            while (t) {
                v1.push_back(t->val);
                t = t->next;
            }
        }
        sort(v1.begin(),v1.end());
        for (int i = 0; i < v1.size(); ++i) { 
                cur->next = new ListNode(v1[i]);
                cur = cur->next;
        }
        return dummy->next;
    }
};

从别人的基数排序那里获得的启发

方法二：基数排序

也学习一下方法

class Solution {
public:
    ListNode* mergeKLists(vector& lists) {
        ListNode *dummy = new ListNode(-1), *cur = dummy;
        unordered_map m;
        int mx = INT_MIN, mn = INT_MAX;
        for (auto node : lists) {
            ListNode *t = node;
            while (t) {
                mx = max(mx, t->val);
                mn = min(mn, t->val);
                ++m[t->val];
                t = t->next;
            }
        }
        for (int i = mn; i <= mx; ++i) {
            if (!m.count(i)) continue;
            for (int j = 0; j < m[i]; ++j) {
                cur->next = new ListNode(i);
                cur = cur->next;
            }
        }
        return dummy->next;
    }
};

方法三：分治 最后两两对比 （重要！）

其中有些细节需要注意 k 和 n 的更新 等等

class Solution {
public:
    ListNode* mergeKLists(vector& lists) {
        if (lists.empty()) return NULL;
        int n = lists.size();
        while (n > 1) {
            int k = (n + 1) / 2;
            for (int i = 0; i < n / 2; ++i) {
                lists[i] = mergeTwoLists(lists[i], lists[i + k]);
            }
            n = k;
        }
        return lists[0];
    }
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode *dummy = new ListNode(-1), *cur = dummy;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                cur->next = l1;
                l1 = l1->next;
            } else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        if (l1) cur->next = l1;
        if (l2) cur->next = l2;
        return dummy->next;
    }
};