【C语言刷LeetCode】92. 反转链表 II(M)

反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

说明:
1 ≤ m ≤ n ≤ 链表长度。

示例:

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reverse-linked-list-ii
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

双指针处理链表问题。

struct ListNode* reverseBetween(struct ListNode* head, int m, int n){
    struct ListNode* pre = NULL;
    struct ListNode* one = head;
    struct ListNode* two;
    struct ListNode* curp;
    struct ListNode* next;
    int i;
    int j;
    int num = n - m + 1;

    if (m == n) return head;

    for(i = 1; i < m; i++) {
        pre = one;
        one = one->next;
    }
    
    two = one;
    
    while(num > 0) {
        next = one->next;
        one->next = curp;
        curp = one;
        one = next;
        num--;
        printf("curp->val=%d,num=%d\n",curp->val, num);
    }

    two->next = one;
    
    if(pre != NULL) {
        pre->next = curp;
        return head;
    } else {
        return curp;
    } 
}
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