71. Merge k Sorted Lists

Merge k Sorted Lists

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

两个方法：方法1. 利用 STL 中的 multiset (根据结点内的值)自动对指针排序。空间 O(N), 时间 O(NlogN).

(亦可用于 k 个无序链表)。（AC: 164ms）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
bool compare (const ListNode *p1, const ListNode *p2) {
     return p1->val < p2->val;
 }
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        typedef bool (*cmp) (const ListNode *p1, const ListNode *p2);
        multiset<ListNode*, cmp> container(compare);
        for(size_t i = 0; i < lists.size(); ++i) {
            ListNode *p = lists[i];
            while(p) { container.insert(p); p = p->next; }
        }
        ListNode *head = NULL, *p = head;
        for(auto it = container.begin(); it != container.end(); ++it) {
            if(p) { p->next = *it; p = *it; } 
            else p = head = *it;
        }
        if(p) p->next = NULL;
        return head;
    }
};

方法2. 不利用任何 STL 函数。对指针建堆排序，只需要一个(win32: 4Byte)指针数组即可。空间 : O(K), 时间 O(NlogK)

(AC: 140ms)

最初代码：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *mergeKLists(vector<ListNode *> &lists) {
12         heap = vector<ListNode*>(lists.size(), 0);
13         sz = 0;
14         for(size_t i = 0; i < lists.size(); ++i) 
15             if(lists[i]) push(lists[i]);
16         ListNode *head = NULL, *p = head;
17         while(sz) {
18             if(head == NULL) 
19                 p = head = pop();
20             else { 
21                 p->next = pop();
22                 p = p->next;
23             }
24             if(p->next) push(p->next);
25         }
26         return head;
27     }
28     void push(ListNode *p) {
29         int child = sz++;
30         while(child > 0) {
31             int father = (child-1) / 2;
32             if(p->val >= heap[father]->val) break;
33             heap[child] = heap[father];
34             child = father;
35         }
36         heap[child] = p;
37      }
38      ListNode* pop() {
39          ListNode *pAns = heap[0];
40          heap[0] = heap[--sz];
41          int father = 0, child = 1;
42          ListNode *p = heap[father];
43          while(child < sz) {
44              if(child+1 < sz && heap[child]->val > heap[child+1]->val) ++child;
45              if(heap[child]->val >= p->val) break;
46              heap[father] = heap[child];
47              father = child;
48              child = 2 * father + 1;
49          }
50          heap[father] = p;
51          return pAns;
52      }
53 private:
54     int sz;
55     vector<ListNode*> heap;
56 };

View Code

优化后（增强易读性）：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Heap {
public:
  Heap(size_t n) : sz(0), heap(vector<ListNode*>(n, NULL)) {}
  void push(ListNode *p);
  ListNode* pop();
  int size() { return sz; }
private:
  int sz;
  vector<ListNode*> heap;
};
inline void Heap::push(ListNode *p) {
    int child = sz++;
    while(child > 0) {
        int father = (child-1) / 2;
        if(p->val >= heap[father]->val) break;
        heap[child] = heap[father];
        child = father;
    }
    heap[child] = p;
}
inline ListNode* Heap::pop() {
    ListNode *pAns = heap[0];
    heap[0] = heap[--sz];
    int father = 0, child = 1;
    ListNode *p = heap[father];
    while(child < sz) {
        if(child+1 < sz && heap[child]->val > heap[child+1]->val) ++child;
        if(heap[child]->val >= p->val) break;
        heap[father] = heap[child];
        father = child;
        child = 2 * father + 1;
    }
    heap[father] = p;
    return pAns;
}
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        Heap heap(lists.size());
        for(size_t i = 0; i < lists.size(); ++i) 
            if(lists[i]) heap.push(lists[i]);
        ListNode *head = NULL, *p = head;
        while(heap.size()) {
            if(head == NULL) 
                p = head = heap.pop();
            else { 
                p->next = heap.pop();
                p = p->next;
            }
            if(p->next) heap.push(p->next);
        }
        return head;
    }
};