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025-Reverse-Nodes-in-k-Group.js
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025-Reverse-Nodes-in-k-Group.js
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/**
* https://leetcode.com/problems/reverse-nodes-in-k-group/description/
* Difficulty:Hard
*
* Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
* k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
* You may not alter the values in the nodes, only nodes itself may be changed.
* Only constant memory is allowed.
* For example,
* Given this linked list: 1->2->3->4->5
* For k = 2, you should return: 2->1->4->3->5
* For k = 3, you should return: 3->2->1->4->5
*
*/
// Definition for singly-linked list.
function ListNode(val) {
this.val = val;
this.next = null;
}
/**
* @param {ListNode} head
* @param {number} k
* @return {ListNode}
*/
var reverseKGroup = function (head, k) {
// if (k === 1)
// return head;
var t = new ListNode(0);
t.next = head;
var s = t;
while (true) {
var cnt = 0;
var f = t;
while (cnt++ < k && f) {
f = f.next;
}
// console.log(p(t), p(f));
if (!f || cnt !== k + 1) break;
cnt = 0;
var a = t.next;
while (++cnt < k) {
var b = a.next;
a.next = b.next;
b.next = t.next;
t.next = b;
// console.log(p(t), p(a), p(b));
}
t = a;
}
return s.next;
};
function p(n) {
var t = n;
var s = '';
while (t) {
s = s + t.val + '->';
t = t.next;
}
s += 'null';
return s;
}
//
console.log(p(reverseKGroup({
val: 1,
next: {
val: 2,
next: {
val: 3,
next: {
val: 4
}
}
}
}, 2)))
console.log(p(reverseKGroup({val: 1}, 2)));
console.log(p(reverseKGroup({
val: 1,
next: {
val: 2
}
}, 2)))
console.log(p(reverseKGroup({
val: 1,
next: {
val: 2,
next: {
val: 3,
next: {
val: 4,
next: {
val: 5,
next: {
val: 6,
next: {
val: 7
}
}
}
}
}
}
}, 3)))
//
console.log(p(reverseKGroup({
val: 1,
next: {
val: 2,
next: {
val: 3,
next: {
val: 4,
next: null
}
}
}
}, 2)));