basic method:快慢指针法
标注解答:
思路分析:
使用快慢双指针:
快指针fast,先走k步,在这里做一个解释:
while(k-- > 0) fast = fast.next;
相当于:
while(k > 0){
fast = fast.next;
k--;
}//走三步:先走再减
如果:
while(--k> 0){
fast = fast.next;
}//走两步:先减再走
class Solution {
public ListNode getKthFromEnd(ListNode head, int k) {
ListNode slow = head, fast = head;
while (k-- > 0) fast = fast.next;
while (fast != null) {
slow = slow.next;
fast = fast.next;
}
return slow;
}
}
作者：AC_OIer
链接：https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/solution/gong-shui-san-xie-yi-ti-san-jie-zhan-dui-w3rz/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
我的解答:
class Solution {
public ListNode getKthFromEnd(ListNode head, int k) {
ListNode pre = head;
ListNode cur = head;
while(cur != null){
if(k > 0){
cur = cur.next;
k--;
}
else{
pre = pre.next;
cur = cur.next;
}
}
return pre;
}
}
