LeetCode题解：《23.合并K个升序链表》

#include <iostream>
#include <vector>
#include "leetcode.h"

using namespace std;

class Solution {
public:
    void sort(vector<ListNode*>& lists) {
        ushort i = 0, length = lists.size();
        for (ushort ii = 0; ii < length; ++ii) {
            if (lists.at(ii) == nullptr)
                continue;
            if (lists.at(i) == nullptr || lists.at(i)->val > lists.at(ii)->val)
                i = ii;
        }
        ListNode* temp = lists.at(i);
        lists.at(i) = lists.at(0);
        lists.at(0) = temp;
    }

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.size() == 0)
            return nullptr;
        sort(lists);
        ListNode* ans = lists.front();

        ListNode* begin, * end, * temp;
        // 如果要归并的链到达尾节点，则跳出循环
        bool flag = true;
        for (int i = 1; i < lists.size(); ++i) {
            begin = ans;
            temp = lists.at(i);
            // 如果该链为空直接跳过
            if (!temp) continue;
            if (nullptr == begin) { ans = temp; continue; }
            while (flag) {
                // 循环找出适合ans链接到temp的node
                while (begin->next && begin->next->val < temp->val) {
                    begin = begin->next;
                }
                // 留一个指针继续搜索ans找到适合temp链接ans的node
                end = begin->next;
                if (nullptr == end) {
                    begin->next = temp;
                    flag = false;
                    continue;
                }

                begin->next = temp;
                ListNode* btemp = temp;
                while (nullptr != temp && nullptr != end && temp->val <= end->val) {
                    btemp = temp;
                    temp = temp->next;
                }
                if (nullptr == temp)
                    flag = false;
                btemp->next = end;
            }
            flag = true;
        }
        return ans;
    }
};