23. Merge k Sorted Lists（看）二分归并（未）

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https://algorithm.yuanbin.me/zh-hans/linked_list/merge_k_sorted_lists.html
解法一，选择归并

/**
 * 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.empty()) return NULL;
        ListNode * dummy = new ListNode(INT_MAX);
        ListNode * lastnode = dummy;
        while(true){
            int count = 0;
            int index = -1, tmp = INT_MAX;
            int i;
            for( i = 0; i < lists.size(); i++){
                if(lists[i] == NULL){                
                    count++;
                    if(count == lists.size()){
                        lastnode->next = NULL;
                        return dummy->next;
                    }
                    continue;
                }
                if(lists[i] != NULL && lists[i]->val < tmp){
                    index = i;
                    tmp = lists[i]->val;
                } 
            }
            lastnode->next = lists[index];
            lastnode = lastnode->next;
            lists[index] = lists[index]->next;
        }
    }
};

解法二：两个的合并，小的逐渐并入！！！要注意合并两个链表的模板

/**
 * 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.empty()) return NULL;
        ListNode * head = lists[0];
        for(int i = 1; i < lists.size();i++){
            head = merge(head, lists[i]);
        }
        return head;
    }
private:
    ListNode* merge(ListNode * left, ListNode * right){
        ListNode * dummy = new ListNode(0);
        ListNode * p = dummy;
        while(left && right){
            if(left->val < right->val){
                p->next = left;
                p = p->next;
                left = left->next;
            }
            else{
                p->next = right;
                p = p->next;
                right = right->next;
            }
        }
        p->next = left ? left : right;
        return dummy->next;
    }
};

解法三：二分归并，不太懂