C++实现LeetCode(19.移除链表倒数第N个节点)

c++ leetcode
这篇文章主要介绍了C++实现LeetCode(19.移除链表倒数第N个节点),本篇文章通过简要的案例,讲解了该项技术的了解与使用,以下就是详细内容,需要的朋友可以参考下

[LeetCode] 19. Remove Nth Node From End of List 移除链表倒数第N个节点

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

这道题让我们移除链表倒数第N个节点,限定n一定是有效的,即n不会大于链表中的元素总数。还有题目要求一次遍历解决问题,那么就得想些比较巧妙的方法了。比如首先要考虑的时,如何找到倒数第N个节点,由于只允许一次遍历,所以不能用一次完整的遍历来统计链表中元素的个数,而是遍历到对应位置就应该移除了。那么就需要用两个指针来帮助解题,pre 和 cur 指针。首先 cur 指针先向前走N步,如果此时 cur 指向空,说明N为链表的长度,则需要移除的为首元素,那么此时返回 head->next 即可,如果 cur 存在,再继续往下走,此时 pre 指针也跟着走,直到 cur 为最后一个元素时停止,此时 pre 指向要移除元素的前一个元素,再修改指针跳过需要移除的元素即可,参见代码如下:

方式一:

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (!head->next) return NULL;
        ListNode *pre = head, *cur = head;
        for (int i = 0; i < n; ++i) cur = cur->next;
        if (!cur) return head->next;
        while (cur->next) {
            cur = cur->next;
            pre = pre->next;
        }
        pre->next = pre->next->next;
        return head;
    }
};

方式二:

class Solution {
public:
    int getLength(ListNode* head) {
        int length = 0;
        while (head) {
            ++length;
            head = head->next;
        }
        return length;
    }

    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummy = new ListNode(0, head);
        int length = getLength(head);
        ListNode* cur = dummy;
        for (int i = 1; i < length - n + 1; ++i) {
            cur = cur->next;
        }
        cur->next = cur->next->next;
        ListNode* ans = dummy->next;
        delete dummy;
        return ans;
    }
};