diff --git a/docs/notes/pics/11548084848123.gif b/docs/notes/pics/11548084848123.gif
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diff --git a/docs/notes/剑指 offer 题解.md b/docs/notes/剑指 offer 题解.md
index b30c78c4..e63bcb23 100644
--- a/docs/notes/剑指 offer 题解.md
+++ b/docs/notes/剑指 offer 题解.md
@@ -116,18 +116,8 @@ Output:
以 (2, 3, 1, 0, 2, 5) 为例，遍历到位置 4 时，该位置上的数为 2，但是第 2 个位置上已经有一个 2 的值了，因此可以知道 2 重复：
-
+
-```text
-position-0 : (2,3,1,0,2,5) // 2 <-> 1
- (1,3,2,0,2,5) // 1 <-> 3
- (3,1,2,0,2,5) // 3 <-> 0
- (0,1,2,3,2,5) // already in position
-position-1 : (0,1,2,3,2,5) // already in position
-position-2 : (0,1,2,3,2,5) // already in position
-position-3 : (0,1,2,3,2,5) // already in position
-position-4 : (0,1,2,3,2,5) // nums[i] == nums[nums[i]], exit
-```
```java
public boolean duplicate(int[] nums, int length, int[] duplication) {
