题解 | 链表内指定区间反转
链表内指定区间反转
https://www.nowcoder.com/practice/b58434e200a648c589ca2063f1faf58c
/** * struct ListNode { * int val; * struct ListNode *next; * ListNode(int x) : val(x), next(nullptr) {} * }; */ class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param head ListNode类 * @param m int整型 * @param n int整型 * @return ListNode类 */ ListNode* reverseBetween(ListNode* head, int m, int n) { // write code here struct ListNode* ans = new ListNode(0); ans->next = head; ListNode *p = head, *q = ans, *l = head, *j = ans, *k = head; if (m == n) { return head; } for (int i = 0; i < m-1; i++) { q = p; p = p->next; } j = q;//区间前一位数 for (int i = 0; i <= n-m; i++) { q = q->next; } k = q->next;//区间后一位数 j->next = q; q = p->next; l = q->next; p->next = k; for(int i = 0; i < n-m-1; i++) { q->next = p; p = q; q = l; l = l->next; } q->next = p; return ans->next; } };