查找链表中倒数第N个结点
小于 1 分钟
查找链表中倒数第N个结点
ListNode* findFromEnd(ListNode* list, int n) {
ListNode* fast = list;
ListNode* slow = list;
while(fast && n--) { //先走N步
fast = fast->next;
}
while(fast) { //当fast为空时 slow刚好指在目标元素上
fast = fast->next;
slow = slow->next;
}
return slow;
}