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`@@ -79,7 +79,7 @@ map目的用来存放我们访问过的元素,因为遍历数组的时候,`
`79`	`79`
`80`	`80`	`所以 map中的存储结构为 {key:数据元素,value:数组元素对应的下标}。`
`81`	`81`
`82`		`-在遍历数组的时候,只需要向map去查询是否有和目前遍历元素比配的数值,如果有,就找到的匹配对,如果没有,就把目前遍历的元素放进map中,因为map存放的就是我们访问过的元素。`
	`82`	`+在遍历数组的时候,只需要向map去查询是否有和目前遍历元素匹配的数值,如果有,就找到的匹配对,如果没有,就把目前遍历的元素放进map中,因为map存放的就是我们访问过的元素。`
`83`	`83`
`84`	`84`	`过程如下:`
`85`	`85`

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`@@ -682,7 +682,33 @@ public class LinkNumbers`
`682`	`682`
`683`	`683`
`684`	`684`
	`685`	`+## 使用虚拟头结点解决链表翻转`
	`686`	`+`
	`687`	`+> 使用虚拟头结点,通过头插法实现链表的翻转(不需要栈)`
	`688`	`+`
	`689`	+```java
	`690`	`+// 迭代方法:增加虚头结点,使用头插法实现链表翻转`
	`691`	`+public static ListNode reverseList1(ListNode head) {`
	`692`	`+ // 创建虚头结点`
	`693`	`+ ListNode dumpyHead = new ListNode(-1);`
	`694`	`+ dumpyHead.next = null;`
	`695`	`+ // 遍历所有节点`
	`696`	`+ ListNode cur = head;`
	`697`	`+ while(cur != null){`
	`698`	`+ ListNode temp = cur.next;`
	`699`	`+ // 头插法`
	`700`	`+ cur.next = dumpyHead.next;`
	`701`	`+ dumpyHead.next = cur;`
	`702`	`+ cur = temp;`
	`703`	`+ }`
	`704`	`+ return dumpyHead.next;`
	`705`	`+}`
	`706`	+```
	`707`	`+`
	`708`	`+`
	`709`	`+`
`685`	`710`	`## 使用栈解决反转链表的问题`
	`711`	`+`
`686`	`712`	`* 首先将所有的结点入栈`
`687`	`713`	`* 然后创建一个虚拟虚拟头结点,让cur指向虚拟头结点。然后开始循环出栈,每出来一个元素,就把它加入到以虚拟头结点为头结点的链表当中,最后返回即可。`
`688`	`714`
`@@ -720,3 +746,4 @@ public ListNode reverseList(ListNode head) {`
`720`	`746`	`<a href="https://programmercarl.com/other/kstar.html" target="_blank">`
`721`	`747`	`<img src="../pics/网站星球宣传海报.jpg" width="1000"/>`
`722`	`748`	`</a>`
	`749`	`+`

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`@@ -476,6 +476,7 @@ class Solution {`
`476`	`476`	`## Python`
`477`	`477`
`478`	`478`	`> 递归法`
	`479`	`+> 常量空间,递归产生的栈不算`
`479`	`480`
`480`	`481`	```python
`481`	`482`	`# Definition for a binary tree node.`
`@@ -521,7 +522,9 @@ class Solution:`
`521`	`522`	```
`522`	`523`
`523`	`524`
`524`		`-> 迭代法-中序遍历-不使用额外空间,利用二叉搜索树特性`
	`525`	`+> 迭代法-中序遍历`
	`526`	`+> 利用二叉搜索树特性,在历遍过程中更新结果,一次历遍`
	`527`	`+> 但需要使用额外空间存储历遍的节点`
`525`	`528`	```python
`526`	`529`	`class Solution:`
`527`	`530`	`def findMode(self, root: TreeNode) -> List[int]:`

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