剑指Offer 22 链表中倒数第k个节点[双指针] HERODING的LeetCode之路

Posted 2021-09-17 HERODING23

解题思路：
第一个方法就是最常规的，遍历一遍找长度，再遍历一遍找倒数第k个，代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* getKthFromEnd(ListNode* head, int k) {
        int len = 0;
        ListNode * demo = head;
        while(demo != nullptr) {
            demo = demo -> next;
            len ++;
        }
        int i = len - k;
        while(i --) {
            head = head -> next;
        }
        return head;
    }
};

快慢指针的方法是更为精妙的，让快指针永远比慢指针快k个节点，那么快指针到末尾的时候，慢指针刚好到倒数第k个，代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* getKthFromEnd(ListNode* head, int k) {
        ListNode* fast = head;
        ListNode* slow = head;
        int i = k;
        while(k --) {
            fast = fast -> next;
        }
        while(fast != nullptr) {
            fast = fast -> next;
            slow = slow -> next;
        }
        return slow;
    }
};