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19. Remove Nth Node From End of List ✅

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

Example 1:

img

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]

Constraints:

  • The number of nodes in the list is sz.
  • 1 <= sz <= 30
  • 0 <= Node.val <= 100
  • 1 <= n <= sz

Follow up: Could you do this in one pass?

Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode cur = head;

        int count = 0;

        while(cur != null){
            cur = cur.next;
            count++;
        }
        /*
        0 1 2 3 4
        1 2 3 4 5
            p
              c
        4 - 2 + 1 = 3
        */
        if (count == 1){
            return null;
        }



        int remove = count - n; // 2 -2 

        if (remove == 0){
            return head.next;
        }
        cur = head;
        ListNode prev = head;

        for (int i = 0; i < remove; i++){
            prev = cur;
            cur = cur.next;
        }

        prev.next = cur.next;
        return head;
    }
}

// TC: O(n)
// SC: O(1)

逻辑没有那么难， special case需要注意

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        // base case
        if (head.next == null){
            return null;
        }


        ListNode dummy = head;

        int count = 0;
        while(dummy != null){
            dummy = dummy.next;
            count++;
        }

        int m = count - n; // 5-2 = 3

        if (m == 0){
            return head.next;
        }
        ListNode cur = head;
        ListNode next = null;
        ListNode prev = null;
        for (int i = 0; i < m; i++){
            prev = cur;
            cur = cur.next; 
        }

        prev.next = cur.next;

        return head;


    }
}
// TC: O(n)
// SC: O(1)

Linked List 和 Tree 似乎都围绕着遍历