LeetCode | 19. 删除链表倒数第 N 个结点

OJ链接

在这里插入图片描述

思路:

struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    //定义虚拟头节点dummy初始化使其指向head
    struct ListNode* dummy = malloc(sizeof(struct ListNode));
    dummy->val = 0;
    dummy->next = head;
    //定义 fast slow 双指针
    struct ListNode* fast = head;
    struct ListNode* slow = dummy;

    for (int i = 0; i < n; ++i) {
        fast = fast-&gt;next;
    }
    while (fast) {
        fast = fast-&gt;next;
        slow = slow-&gt;next;
    }
    slow-&gt;next = slow-&gt;next-&gt;next;//删除倒数第n个节点
    head = dummy-&gt;next;
    free(dummy);//删除虚拟节点dummy
    return head;
}

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