LeetCode题解之Split Linked List in Parts

1、题目描述

2、题目分析

主要是理解题意，将每个子链表应该分得的节点个数计算清楚。利用除数和余数的方法进行计算。

3、代码

 1 vector<ListNode*> splitListToParts(ListNode* root, int k) {
 2         vector<ListNode*> res(k, NULL);
 3         if (root == NULL) {
 4             return res;
 5         }
 6         
 7         int listlen = 0;
 8         ListNode *p = root;
 9         while (p != NULL) {
10             listlen++;
11             p = p->next;
12         }
13         
14         int m = listlen / k;
15         int n = listlen % k;
16         
17         if (m == 0) {
18             p = root;
19             for (int i = 0; i < n; i++) {
20                 res[i] = p;
21                 ListNode *pn = p->next;
22                 p->next = NULL;
23                 p = pn;
24             }
25         } else {
26             p = root;
27             int i;
28             for(i = 0; i < k; i++) {
29                 ListNode *pn;
30                 res[i] = p;
31                 int step = (i < n)? m:m-1;
32                 while(step--) {
33                     p = p->next;
34                 }
35                 pn = p->next;
36                 p->next = NULL;
37                 p = pn;
38                 if (p == NULL)
39                     break;
40                 }
41             }
42         
43         
44         return res;
45         
46         
47                
48     }