Summary on LinkedList reversing.
Starter
Special case: LC206 reverse all the nodes in one pass:
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public ListNode reverseList(ListNode head) {
ListNode newHead = null;
while (head != null) {
ListNode next = head.next;
head.next = newHead;
newHead = head;
head = next;
}
return newHead;
}
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General case: LC92 reverse nodes from position m to n
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public ListNode reverseBetween(ListNode head, int m, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode pre = dummy, end, cur;
for (int i = 0; i < m-1; i++) { pre = pre.next; }
end = pre.next;
cur = end.next;
for (int i = 0; i < n - m; i++) {
end.next = cur.next;
cur.next = pre.next;
pre.next = cur;
cur = end.next;
}
return dummy.next;
}
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Main dish: LC25 Reverse Nodes in k-Groups
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/**
* Reverse a link list between pre and next exclusively
* e.g.:
* 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 precedence of parameter next
*/
private static ListNode reverse(ListNode pre, ListNode next) {
ListNode end = pre.next, cur = end.next;
while (cur != next) {
end.next = cur.next;
cur.next = pre.next;
pre.next = cur;
cur = end.next;
}
return end;
}
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null || k == 1) return head;
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode pre = dummy;
int i = 0;
while (head != null) {
head = head.next;
if (++i % k == 0) pre = reverse(pre, head);
}
return dummy.next;
}
/**
* Note: the while loop is the same as following process, which is easier to understand:
* while(head != null){
* i++;
* if(i % k ==0){
* pre = reverse(pre, head.next);
* head = pre.next;
* }else {
* head = head.next;
* }
* }
*/
}
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