[LeetCode] 876. Middle of the Linked List

Given a non-empty, singly linked list with head node head, return a middle node of linked list.
If there are two middle nodes, return the second middle node.
Example 1:
Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.
Note:
The number of nodes in the given list will be between 1 and 100.
If there are two middle nodes, return the second middle node.

方法1: 先取长度再求中点

let middleNode = function(head) {
    let targetLen = Math.ceil(getLength(head)/2);
    return returnAtPosition(head, targetLen);
};

let getLength = function(node) {
    let length = 0;
    while(node.next) {
        length += 1;
        node = node.next;
    }
    return length;
};

let returnAtPosition = function(node, targetLen) {
    for(let i = 0; i < targetLen; i++) {
        node = node.next;
    }
    return node;
};

方法2：快慢指针

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var middleNode = function(head) {
  if (head === null) {
      return false; 
  }
    
  var slow = head;
  var fast = head;
  while (true) {
      if (fast.next === null) {
          return slow;
      }
      
      if (fast.next.next === null) {
          return slow.next;
      }
      
      fast = fast.next.next
      slow = slow.next
  }
};