自己寫的太復雜了，一開始想的是給開始的lists頭們排序，然后從這個序列的第一個抽出來，然后再重新用二分法進行排序，不過這個方法large超時了，看了網上的發現還是用很土地方法用一個for循環從前兩個開始merge到最后，不知道為什么自己把這個想這么復雜。
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 *merge(ListNode *p, ListNode *q) { 12 if (!p) return q; 13 if (!q) return p; 14 ListNode *ret = NULL; 15 ListNode *reti; 16 while (p && q) { 17 if (!ret) { 18 ret = new ListNode(min(p->val, q->val)); 19 reti = ret; 20 if (p->val < q->val) p = p->next; 21 else q = q->next; 22 } 23 else { 24 ListNode *tmp = new ListNode(min(p->val, q->val)); 25 ret->next = tmp; 26 ret = ret->next; 27 if (p->val < q->val) p = p->next; 28 else q = q->next; 29 } 30 } 31 if (!p) ret->next = q; 32 if (!q) ret->next = p; 33 return reti; 34 } 35 ListNode *mergeKLists(vector<ListNode *> &lists) { 36 // Start typing your C/C++ solution below 37 // DO NOT write int main() function 38 if (!lists.size()) return NULL; 39 ListNode *ret = lists[0]; 40 for (int i = 1; i < lists.size(); i++) ret = merge(ret, lists[i]); 41 return ret; 42 } 43 };
?推薦下面這個更符合面試的代碼
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 struct cmp { 10 bool operator()(ListNode *a, ListNode *b) { 11 return a->val > b->val; 12 } 13 }; 14 class Solution { 15 public: 16 ListNode *mergeKLists(vector<ListNode *> &lists) { 17 priority_queue<ListNode*, vector<ListNode*>, cmp> S; 18 if (lists.size() == 0) return NULL; 19 for (int i = 0; i < lists.size(); i++) { 20 if (lists[i]) S.push(lists[i]); 21 } 22 ListNode *ans = NULL; 23 ListNode *p = ans; 24 while (!S.empty()) { 25 ListNode *tmp = S.top(); 26 if (!ans) { 27 ans = tmp; 28 p = ans; 29 } 30 else { 31 p->next = tmp; 32 p = p->next; 33 } 34 if (!tmp->next) S.pop(); 35 else { 36 tmp = tmp->next; 37 S.pop(); 38 S.push(tmp); 39 } 40 } 41 return ans; 42 } 43 };
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