链表中点取值
其实很好弄，条件是fast.next!=null,fast.next.next!=null
画四个点，把两个指针的位置画一下根据条件判断slow的落点就行了
1、偶数上中点，基数中点
public static Node midOrUpMidNode(Node head) {
if (head == null || head.next == null || head.next.next == null) {
return head;
}
// 链表有3个点或以上
Node slow = head.next;
Node fast = head.next.next;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
2、偶数下中点，基数中点
public static Node midOrDownMidNode(Node head) {
if (head == null || head.next == null) {
return head;
}
Node slow = head.next;
Node fast = head.next;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
3、偶数上中点前一个，基数中点前一个
public static Node midOrUpMidPreNode(Node head) {
if (head == null || head.next == null || head.next.next == null) {
return null;
}
Node slow = head;
Node fast = head.next.next;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
4、偶数下中点前一个，基数中点前一个（允许存在两位，那么就是头节点）
public static Node midOrDownMidPreNode(Node head) {
if (head == null || head.next == null) {
return null;
}
if (head.next.next == null) {
return head;
}
Node slow = head;
Node fast = head.next;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
⑨✌们拿它练练面试愣是给他整出幻觉了