add: K个一组翻转链表
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@ -411,3 +411,8 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库,代码中对重
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- LeetCode 876. 链表的中间结点 <https://leetcode-cn.com/problems/middle-of-the-linked-list/>
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- LintCode 1609. 链表的中间结点 <https://www.lintcode.com/problem/middle-of-the-linked-list/description>
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- [K个一组翻转链表](src/list/reverse-nodes-in-k-group.js)
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- LeetCode 25. K 个一组翻转链表 <https://leetcode-cn.com/problems/reverse-nodes-in-k-group/>
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- LintCode 450. K组翻转链表 <https://www.lintcode.com/problem/reverse-nodes-in-k-group/>
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49
src/list/reverse-nodes-in-k-group.js
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49
src/list/reverse-nodes-in-k-group.js
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@ -0,0 +1,49 @@
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} head
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* @param {number} k
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* @return {ListNode}
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*/
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export const reverseKGroup = function (head, k) {
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let sum = 0
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let start = head
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let newStart = head
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let first = true
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const last = []
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while (head) {
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if (++sum === k) {
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const headNext = head.next
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last.push(start)
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let next = head.next
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for (let i = 0; i < sum - 1; i++) {
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const tmp = start.next
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start.next = next
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next = start
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start = tmp
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}
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start.next = next
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if (first) {
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newStart = start
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first = false
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} else {
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const la = last.shift()
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la.next = head
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}
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sum = 0
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start = headNext
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head = headNext
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} else {
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head = head.next
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}
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}
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return newStart
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}
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@ -15,3 +15,13 @@ export const arrToList = (arr) => {
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return head
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}
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export const listToArr = (list) => {
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const arr = []
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while (list) {
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arr.push(list.val)
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list = list.next
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}
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return arr
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}
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11
test/list/reverse-nodes-in-k-group.test.js
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11
test/list/reverse-nodes-in-k-group.test.js
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@ -0,0 +1,11 @@
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import { reverseKGroup } from '../../src/list/reverse-nodes-in-k-group'
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import { arrToList, listToArr } from './ListNode'
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test('K个一组翻转链表', () => {
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const res = reverseKGroup(arrToList([1, 2, 3, 4, 5]), 3)
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expect(listToArr(res)).toEqual([3, 2, 1, 4, 5])
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expect(res).toEqual(arrToList([3, 2, 1, 4, 5]))
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expect(reverseKGroup(arrToList([1, 2, 3, 4, 5]), 2)).toEqual(arrToList([2, 1, 4, 3, 5]))
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expect(reverseKGroup(arrToList([1, 2, 3, 4, 5, 6]), 2)).toEqual(arrToList([2, 1, 4, 3, 6, 5]))
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expect(reverseKGroup(arrToList([1, 2, 3, 4, 5, 6, 7, 8]), 2)).toEqual(arrToList([2, 1, 4, 3, 6, 5, 8, 7]))
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})
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