30-Day Leetcoding Challenge Day8

两种方法：时间复杂度都为O(N)

1.转为数组保存，求出总长度的一半 空间复杂度O(n)

2.快慢指针法，输出慢指针 空间复杂度O(1)

JAVA

class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode[] res = new ListNode[100];
        int i = 0;
        while(head != null){
            res[i++] = head;
            head = head.next;
        }
        return res[i/2];
    }
}

class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}

Python3

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        res = []
        while head:
            res.append(head)
            head = head.next
        return res[len(res)//2]

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        slow = head
        fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow