Comments (1)
纯链表加索引解决的思路:
拆分为2个函数
只反转链表
在找到需要反转的链表,用 1 去反转
每一轮循环中,将链表视为
pre => start => *** => end => next
我们翻转 start => *** => end 这一段, 变为
pre end => *** => start next
再将原链表和翻转后的合并即可,然后寻找下一段要反转的链表继续循环
pre => end => *** => start => next
// 定义链表节点
class Node {
constructor(v, last) {
this.next = null
this.value = v
if (last) {
last.next = this
}
}
}
let head = makeList(10)
console.info(toString(head))
head = invert(head, 3)
console.info(toString(head))
// 原链表: 0,1,2,3,4,5,6,7,8,9
// 反转后: 2,1,0,5,4,3,8,7,6,9
// ---- 核心** -----
/**
* 链表每m个节点反转
* @param {*} head
* @param {*} n
*/
function invert(head, n) {
let pre = new Node(999, null)
pre.next = head
let start = head
let end = head
// 缓存头
head = pre
let count = 1
while (start && end) {
// 查找需要反转的链表 end 节点
if (count < n) {
count++
end = end.next
} else {
count = 1
// 缓存对 end 之后的链表的索引
let next = end.next
// 反转 start -> ** -> end 这段链表
revert(start, next)
// 反转成功后, end 节点是新链表的头了, 而 start 是尾了
pre.next = end
// 反转的链表连上剩下的链表
start.next = next
// 初始化下一段要反转的链表段
pre = start
start = next
end = next
}
}
return head.next
}
/**
* 给定一个链表头和结束标记进行反转
* @param {*} start
* @param {*} endTag
*/
function revert(start, endTag) {
let temp
let pre = null
while (start !== endTag) {
temp = start.next
start.next = pre
pre = start
start = temp
}
return pre
}
// ----工具-----
// 构造一个链表
function makeList(length) {
const head = new Node(-1, null)
Array.from({length}, (_, i) => i).reduce((last, key) => new Node(key, last), head)
return head.next
}
// 打印链表
function toString(head) {
let temp = new Node(999, null)
temp.next = head
let show = []
while ((temp = temp.next)) {
show.push(temp.value)
}
return show.join(',')
}
from interview.
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from interview.