剑指 Offer 22. 链表中倒数第 K 个节点 - Leetcode

解题思路：

第一时间想到的解法：

先遍历统计链表长度，记为 $n$ ；
设置一个指针走 $(n-k)$ 步，即可找到链表倒数第 $k$ 个节点；

使用双指针则可以不用统计链表长度。

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算法流程：

初始化： 前指针 former 、后指针 latter ，双指针都指向头节点 head 。
构建双指针距离： 前指针 former 先向前走 $k$ 步（结束后，双指针 former 和 latter 间相距 $k$ 步）。
双指针共同移动： 循环中，双指针 former 和 latter 每轮都向前走一步，直至 former 走过链表 尾节点 时跳出（跳出后， latter 与尾节点距离为 $k-1$，即 latter 指向倒数第 $k$ 个节点）。
返回值： 返回 latter 即可。

<,,,,,,,>

复杂度分析：

时间复杂度 $O(N)$ ： $N$ 为链表长度；总体看， former 走了 $N$ 步， latter 走了 $(N-k)$ 步。
空间复杂度 $O(1)$ ： 双指针 former , latter 使用常数大小的额外空间。

代码：

Python

class Solution:
    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
        former, latter = head, head
        for _ in range(k):
            former = former.next
        while former:
            former, latter = former.next, latter.next
        return latter

Java

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode former = head, latter = head;
        for(int i = 0; i < k; i++)
            former = former.next;
        while(former != null) {
            former = former.next;
            latter = latter.next;
        }
        return latter;
    }
}

C++

class Solution {
public:
    ListNode* getKthFromEnd(ListNode* head, int k) {
        ListNode *former = head, *latter = head;
        for(int i = 0; i < k; i++)
            former = former->next;
        while(former != nullptr) {
            former = former->next;
            latter = latter->next;
        }
        return latter;
    }
};

本题没有 $k>$ 链表长度的测试样例，因此不用考虑越界。考虑越界问题的代码如下：

Python

class Solution:
    def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
        former, latter = head, head
        for _ in range(k):
            if not former: return
            former = former.next
        while former:
            former, latter = former.next, latter.next
        return latter

Java

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode former = head, latter = head;
        for(int i = 0; i < k; i++) {
            if(former == null) return null;
            former = former.next;
        }
        while(former != null) {
            former = former.next;
            latter = latter.next;
        }
        return latter;
    }
}

C++

class Solution {
public:
    ListNode* getKthFromEnd(ListNode* head, int k) {
        ListNode *former = head, *latter = head;
        for(int i = 0; i < k; i++) {
            if(former == nullptr) return nullptr;
            former = former->next;
        }
        while(former != nullptr) {
            former = former->next;
            latter = latter->next;
        }
        return latter;
    }
};