Remove Nth Node From End of List

最新推荐文章于 2024-06-20 12:29:29 发布

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最新推荐文章于 2024-06-20 12:29:29 发布

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分类专栏： Leetcode

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直观思维先走到底得出node数目count，再从前走count - N。很明显时间复杂度2n，这是最直白最普通的思维，没有联系计算机相关数据结构知识，所以效率差。之后看了解答用了双指针，让fast先走N步然后slow再同时走，那么fast到底时slow就是目标位置。这才是计算机式的解答。

本人AC版：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *removeNthFromEnd(struct ListNode *head, int n) {
    if (n == 0)
        return head;
    int count = 0;
    struct ListNode *node = head;
    while ((node)) {
        node = node->next;
        count++;
    }
    node = head;
    if (n == count) {
        head = head->next;
        free(node);
    }
    else {
        for (int i = 0; i < count - n - 1; i++)
            node = node->next;
        struct ListNode *tmpnode = node->next;
        node->next = node->next->next;
        free(tmpnode);
    }
    return head;
}

解答nice版：

struct ListNode *fixed_removeNthFromEnd(struct ListNode *head, int n) {
    ListNode dummy = {-1, head};
    ListNode *xp = head, *yp = &dummy;
    for (int i = 1; i < n; i++)
        xp = xp->next;
    
    while (xp->next) {
        yp = yp->next;
        xp = xp->next;
    }
    
    ListNode *tmpp = yp;
    yp->next = yp->next->next;
    free(tmpp->next);
    return dummy.next;
}