Merge pull request #46 from AttackXiaoJinJin/webchen-branch
[JavaScript][07_linkedlist] 完善链表代码
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commit
e6bf08b7c2
@ -33,7 +33,7 @@ class LinkedList {
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currentNode = currentNode.next
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pos++
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}
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return currentNode === null ? -1 : pos
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return currentNode === null ? -1 : currentNode
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}
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// 指定元素向后插入
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insert(newElement, element) {
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@ -69,6 +69,9 @@ class LinkedList {
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}
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// 遍历显示所有节点
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display() {
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//先检查是否为环
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if(this.checkCircle()) return false
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let currentNode = this.head
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while (currentNode !== null) {
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console.log(currentNode.element)
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@ -89,6 +92,30 @@ class LinkedList {
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this.head = root
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}
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//增强尾插法可读性,便于初学者理解
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reverseList1(){
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//head节点即哨兵,作用就是使所有链表,
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// 包括空链表的头节点不为null,并使对单链表的插入、删除操作不需要区分是否为空表或是否在第一个位置进行,
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// 从而与其他位置的插入、删除操作一致
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//所以反转链表的时候不需要带上head节点
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let currentNode=this.head.next
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//第一个节点头结点让其指向null
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let previousNode=null
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while(currentNode!==null){
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//务必先保留下一节点的指针地址
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let nextNode=currentNode.next
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//第一次是null
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currentNode.next=previousNode
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//此时将previousNode赋值为当前节点,
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// 那么下次循环的时候,方便下次的currentNode指向previousNode
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previousNode=currentNode
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//抬走,下一个!
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currentNode=nextNode
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}
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//最后将反转好的链表加上头节点
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this.head.next=previousNode
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}
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// 自己一开始瞎想的。差距啊
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reverseList2() {
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let currentNode = this.head.next
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@ -122,6 +149,8 @@ class LinkedList {
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}
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// 删除倒数第k个节点
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removeByIndexFromEnd(index) {
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//务必先判断是否是 环链表
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if(this.checkCircle()) return false
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let pos = 1
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this.reverseList()
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let currentNode = this.head.next
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