92. 反转链表 II
webVueBlog opened this issue · 2 comments
webVueBlog commented
给你单链表的头指针 head 和两个整数 left 和 right ,其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点,返回 反转后的链表 。
示例 1:
输入:head = [1,2,3,4,5], left = 2, right = 4
输出:[1,4,3,2,5]
示例 2:
输入:head = [5], left = 1, right = 1
输出:[5]
提示:
链表中节点数目为 n
1 <= n <= 500
-500 <= Node.val <= 500
1 <= left <= right <= n
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} left
* @param {number} right
* @return {ListNode}
92. 反转链表 II
left = 2 right = 4
定义一个虚拟的头节点
-1 -> 1 -> 2 -> 3 -> 4 -> 5
pre cur
pre cur
pre cur
2 -> 3 -> 4
p q
*/
var reverseBetween = function(head, left, right) {
const newNode = new ListNode(-1, head)
let pre = newNode, cur = head, next
let n = right - left + 1
while(--left) {
pre = pre.next
cur = cur.next
}
while(--n) {
cur = cur.next
}
next = cur.next
cur.next = null
const reverseHead = reverse(pre.next, next)
pre.next = reverseHead
return newNode.next
};
var reverse = function (head, next) {
let pre = head, cur = head.next
while(head.next) {
head.next = cur.next
cur.next = pre
pre = cur
cur = head.next
}
head.next = next
return pre
}
webVueBlog commented
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} left
* @param {number} right
* @return {ListNode}
92. 反转链表 II
left = 2 right = 4
定义一个虚拟的头节点
*/
var reverseBetween = function(head, left, right) {
const newNode = new ListNode(-1, head)
let pre = newNode, cur = head
let n = right - left + 1
while(--left) {
pre = pre.next
cur = cur.next
}
while(--n) {
const q = cur.next
cur.next = q.next
q.next = pre.next
pre.next = q
}
return newNode.next
};
webVueBlog commented
var reverseBetween = function(head, left, right) {
const dummyNode = new ListNode(-1);
dummyNode.next = head; // 虚拟头节点
let pre = dummyNode;
for (let i = 0; i < left - 1; i++) { // pre遍历到left的前一个节点
pre = pre.next;
}
let rightNode = pre;
for (let i = 0; i < right - left + 1; i++) { // rightNode遍历到right的位置
rightNode = rightNode.next;
}
let leftNode = pre.next; // 保存leftNode
let curr = rightNode.next; // 保存rightNode.next
// 切断left到right的子链
pre.next = null;
rightNode.next = null;
// 反转left到right的子链
reverseLinkedList(leftNode);
// 返乡连接
pre.next = rightNode;
leftNode.next = curr;
return dummyNode.next;
};
const reverseLinkedList = (head) => {
let pre = null;
let cur = head;
while (cur) {
const next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
}