问题描述
Given a list, rotate the list to the right by k places, where k is non-negative.
For example:
Given1->2->3->4->5->NULLand k =2,
return4->5->1->2->3->NULL.
问题分析
这题很麻烦,n可能很大,需要取余才能确定分界点,链表操作没有什么难点。
代码实现
public ListNode rotateRight(ListNode head, int n) {
if (head == null) return null;
ListNode tmp = head;
int len = 0;
while (tmp != null) {
len++;
tmp = tmp.next;
}
n = n % len;
if (n == 0) return head;
ListNode cur, fast, slow;
cur = fast = slow = head;
for (int i = 0; i < n; i++) {
if (fast != null) fast = fast.next;
else return null;
}
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
ListNode newHead = slow.next;
slow.next = null;
fast.next = cur;
return newHead;
}
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