diff --git a/docs/notes/剑指 Offer 题解 - 20~29.md b/docs/notes/剑指 Offer 题解 - 20~29.md
index 51d49793..0480b2c6 100644
--- a/docs/notes/剑指 Offer 题解 - 20~29.md
+++ b/docs/notes/剑指 Offer 题解 - 20~29.md
@@ -68,7 +68,7 @@ public boolean isNumeric(char[] str) {
需要保证奇数和奇数,偶数和偶数之间的相对位置不变,这和书本不太一样。
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## 解题思路
@@ -98,7 +98,7 @@ public void reOrderArray(int[] nums) {
设链表的长度为 N。设两个指针 P1 和 P2,先让 P1 移动 K 个节点,则还有 N - K 个节点可以移动。此时让 P1 和 P2 同时移动,可以知道当 P1 移动到链表结尾时,P2 移动到 N - K 个节点处,该位置就是倒数第 K 个节点。
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```java
public ListNode FindKthToTail(ListNode head, int k) {
@@ -132,8 +132,7 @@ public ListNode FindKthToTail(ListNode head, int k) {
在相遇点,slow 要到环的入口点还需要移动 z 个节点,如果让 fast 重新从头开始移动,并且速度变为每次移动一个节点,那么它到环入口点还需要移动 x 个节点。在上面已经推导出 x=z,因此 fast 和 slow 将在环入口点相遇。
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```java
public ListNode EntryNodeOfLoop(ListNode pHead) {
@@ -194,7 +193,7 @@ public ListNode ReverseList(ListNode head) {
## 题目描述
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## 解题思路
@@ -246,7 +245,7 @@ public ListNode Merge(ListNode list1, ListNode list2) {
## 题目描述
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## 解题思路
@@ -274,7 +273,7 @@ private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
## 题目描述
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## 解题思路
@@ -300,7 +299,7 @@ private void swap(TreeNode root) {
## 题目描述
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## 解题思路
@@ -330,7 +329,7 @@ boolean isSymmetrical(TreeNode t1, TreeNode t2) {
下图的矩阵顺时针打印结果为:1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10
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## 解题思路
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