LeetCode: Reverse Nodes in k-Group 解题报告
Reverse Nodes in k-Group
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

/*
SOLUTION 2: A better rec version.
*/
public ListNode reverseKGroup2(ListNode head, int k) {
if (head == null) {
return null;
} return rec2(head, k);
} public ListNode rec2(ListNode head, int k) {
if (head == null) {
return null;
} ListNode dummy = new ListNode(0); ListNode cur = head;
int cnt = 0;
while (cur != null) {
ListNode tmp = cur.next;
cur.next = dummy.next;
dummy.next = cur; cur = tmp; cnt++; // reverse a k group.
if (cnt == k) {
// BUG 1:
head.next = rec2(tmp, k);
return dummy.next;
}
} // we don't have k nodes.
if (cnt != k) {
cur = dummy.next;
dummy.next = null; // reverse again.
while (cur != null) {
ListNode tmp = cur.next;
cur.next = dummy.next;
dummy.next = cur; cur = tmp;
}
} return dummy.next;
}
SOLUTION 2
另一个思路的递归:
先查看有没有k个node,如果有,切开2个链表,反转当前链表,并且使用递归处理下一个section,最后再把2者连接起来即可。
public ListNode reverseKGroup1(ListNode head, int k) {
if (head == null) {
return null;
} return rec(head, k);
} // Solution 1: Recursion.
public ListNode rec(ListNode head, int k) {
// Reverse k and link to the next section.
ListNode dummy = new ListNode(0);
dummy.next = head; // find the tail node of the section. If not find, just return.
int cnt = k;
ListNode tail = dummy;
while (cnt > 0 && tail != null) {
cnt--;
tail = tail.next;
} // We don't have k nodes to revers.
// bug 1: we should judge that if tail == null to avoid the overflow.
if (tail == null) {
return head;
} // cut the 2 list.
ListNode next = tail.next;
tail.next = null; // reverse the first list.
ListNode newHead = reverse(head); // reverse the next section.
next = rec(next, k); // link the 2 sections.
head.next = next; return newHead;
} public ListNode reverse(ListNode head) {
ListNode dummy = new ListNode(0);
while (head != null) {
ListNode tmp = head.next;
head.next = dummy.next;
dummy.next = head; head = tmp;
} return dummy.next;
}
SOLUTION 3
使用一个专用的反转函数来进行反转,从头到尾遍历,遍历到K的时候,使用Pre-Next指针的方式进行反转。这个方法比递归更棒。
要特别注意的是:
reverseSection 函数中,while 循环的终止条件不是cur != null,而是cur != next。这一点要特别注意,否则很容易造成死循环!
// BUG: Severe. if we use cur != null here, we will cause very serious loop error.
while (cur != next) {
...
}
/*
SOLUTION 3: A Iteration version.
*/
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null) {
return null;
} ListNode dummy = new ListNode(0);
dummy.next = head; ListNode pre = dummy;
ListNode cur = pre.next; int cnt = 0;
while (cur != null) {
cnt++;
cur = cur.next; if (cnt == k) {
cnt = 0;
pre = reverseSection(pre, cur);
cur = pre.next;
}
} return dummy.next;
} /**
* Reverse a link list between pre and next exclusively
* an example:
* a linked list:
* 0->1->2->3->4->5->6
* | |
* pre next
* after call pre = reverse(pre, next)
*
* 0->3->2->1->4->5->6
* | |
* pre next
* @param pre
* @param next
* @return the reversed list's last node, which is the precedence of parameter next
*/
private static ListNode reverseSection(ListNode pre, ListNode next){
ListNode cur = pre.next; // record the new tail.
ListNode tail = cur; // BUG: Severe. if we use cur != null here, we will cause very serious loop error.
while (cur != next) {
ListNode tmp = cur.next;
cur.next = pre.next;
pre.next = cur;
cur = tmp;
} tail.next = next;
return tail;
}
GITHUB:
1. 主页君的GitHub代码
2. 2014.1227 Redo:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/list/ReverseKGroup_1227_2014.java
ref: http://www.cnblogs.com/lichen782/p/leetcode_Reverse_Nodes_in_kGroup.html
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