Merge k Sorted Lists

題目

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

 

思路

K 路歸併鏈表；

注意：不能用 ListNode *cur 來記錄當前的鏈表節點（因爲這樣的話無法修改數組中的鏈表），而是要直接在數組中修改鏈表，所以要用 index 記錄要修改的鏈表。

假設所有的節點總數是 Sum， 共有 Num 個鏈表， 時間複雜度是：O(Num*Sum).

 

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int num = lists.size();
        if(num==0)  return NULL;
        if(num==1)  return lists[0];
        ListNode *head = NULL;
        ListNode *tail = NULL;
        int index = -1;
        while(true) {
            int min = INT_MAX;
            index = -1;
            for(int i=0;i<num;i++) {
                if(lists[i]) { 
                    if(lists[i]->val<min) {
                        min = lists[i]->val;
                        index = i;
                    }
                }
            }
            // There is no extra node to sort
            if(index<0) 
                break;
            if(head==NULL)
                head = lists[index];
            else
                tail->next = lists[index];
            tail = lists[index];
            lists[index] = lists[index]->next;                        
        }
        return head;
    }
};