剑指offer 19.删除链表中倒数第N个节点
解题思路1:
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode fast = head, slow = head;
for(int i = 0; i < n; i++){
fast = fast.next;
}
if(fast == null){
return head.next;
}
while(fast.next != null){
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return head;
}
}
解题思路2
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode tmp = new ListNode(0, head);
int len = 0;
while(tmp.next != null){
len++;
tmp = tmp.next;
}
ListNode node = new ListNode(0, head);
int count = 1;
while(true){
if(n == len){
node.next = head.next;
break;
}else if(count == len - n){
head.next = head.next.next;
break;
}
head = head.next;
count++;
}
return node.next;
}
}
剑指offer 22.链表中倒数第k个节点
解题思路1 快慢指针 让快指针先走k 步, 当fast != null 时, slow 指针刚好指向链表中倒数第k 个节点
class Solution {
public ListNode getKthFromEnd(ListNode head, int k) {
ListNode fast = head, slow = head;
for(int i = 1; i <= k; i++){
fast = fast.next;
}
while(fast != null){
fast = fast.next;
slow = slow.next;
}
return slow;
}
}
解题思路2: 遍历链表, 统计链表的长度count 链表中倒数第k 个节点 = 从头开始遍历count - k + 1 个节点处
class Solution {
public ListNode getKthFromEnd(ListNode head, int k) {
if(head == null && head.next == null){
return head;
}
int count = 0;
ListNode p = head;
while(p.next != null){
count++;
p = p.next;
}
for(int i = 1; i <= count - k + 1; i++){
head = head.next;
}
return head;
}
}
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