题解 | #合并k个已排序的链表#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param lists ListNode类vector
     * @return ListNode类
     */

    ListNode* findMinHead(vector<ListNode*>& lists);
    ListNode* mergeKLists(vector<ListNode*>& lists) {

        // write code here
        //思路,创建链表头节点,其实就是找k的链
        //表的头最小的那个一个,拼接到新建链表的尾部
        ListNode* head = new ListNode(0);
        ListNode* curr = head;

        while (ListNode* ptr = findMinHead(lists)) {
            curr->next = ptr;
            curr = curr->next;


        }
        ListNode* ret = head->next;
        delete head;
        return ret;

    }
};

ListNode* Solution::findMinHead(vector<ListNode*>& lists) {

    ListNode* ret = nullptr;
    int n = 0; //记录需要返回的下标
    for (int i = 0; i < lists.size(); ++i) {

        if (!ret && lists[i]) {
            ret = lists[i];
            n = i;
        } else if (ret && lists[i]) {
            if (lists[i]->val < ret->val) {
                ret = lists[i];
                n = i;
            }
        }
    }
    if (ret) {
        //cout<<ret->val<<endl;
        lists[n] = lists[n]->next;
    }

    return ret;
}