标签：style str nod null init etc 方法 val ++
问题比较简单，方法就是k%sum，然后移动这么多次，把这段链表插在首部就搞定了。有几个测试用例比较恶心，k==0 head==NULL和k%sum==0的情况，假如k%sum为0的话根本不用任何操作直接返回head就好。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* rotateRight(ListNode* head, int k) { if (head == NULL) return NULL; if (k == 0) return head; int sum = 0; ListNode* st; st = head; ListNode* dummyNode; dummyNode = new ListNode(0); dummyNode->next = head; while (st != NULL) { st = st->next; sum++; } if (sum == 1) return head; if (k > sum) { k = k%sum; } int n; n = sum - k; st = dummyNode; while (n--) { st = st->next; } if (k==0) return head; ListNode* p; p= st->next; st->next = NULL; ListNode* q; q = p; while (q!=NULL&&q->next != NULL) { q = q->next; } q->next = dummyNode->next; dummyNode->next = p; return dummyNode->next; } };
标签：style str nod null init etc 方法 val ++
原文地址：https://www.cnblogs.com/legendcong/p/9716548.html
