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diff --git a/_posts/2024-03-04-好文.md b/_posts/2024-03-04-好文.md
@@ -18,7 +18,7 @@ img_path: /assets/
 - 动态规划的解题套路
 - leetcode案例分析
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/01fdad6c74864d7da07193203c2dcbf4~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/a.png)
 
 公众号：**捡田螺的小男孩**
 
@@ -36,7 +36,7 @@ img_path: /assets/
 
 动态规划最核心的思想，就在于**拆分子问题，记住过往，减少重复计算**。
 
-![动态规划在于记住过往](img/2024-03-04-%E5%A5%BD%E6%96%87/d26f967ff6e447d291b0b196c4edaa07~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![动态规划在于记住过往](img/2024-03-04-%E5%A5%BD%E6%96%87/b.png)
 
 我们来看下，网上比较流行的一个例子：
 
@@ -99,11 +99,11 @@ scss
 
 去leetcode提交一下，发现有问题，超出时间限制了
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/3e0608ec89a246568e01c7ba4b8f50d8~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/c.png)
 
 为什么超时了呢？递归耗时在哪里呢？先画出**递归树**看看：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/2dd95840552e4a7db9536884b7a9b558~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/d.png)
 
 - 要计算原问题 f(10)，就需要先计算出子问题 f(9) 和 f(8)
 - 然后要计算 f(9)，又要先算出子问题 f(8) 和 f(7)，以此类推。
@@ -130,19 +130,19 @@ scss
 
 - 第一步，f（10）= f(9) + f(8)，f(9) 和f（8）都需要计算出来，然后再加到备忘录中，如下：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/94484290462e45c799609b0a581b14b0~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/e.png)
 
 - 第二步，  f(9) = f（8）+ f（7），f（8）= f（7）+ f(6), 因为 f(8) 已经在备忘录中啦，所以可以省掉，f(7),f（6）都需要计算出来，加到备忘录中~
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/12900b17d77f4adfbdb71920f729d61a~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/a1.png)
 
 第三步， f(8) = f（7）+ f(6),发现f(8)，f(7),f（6）全部都在备忘录上了，所以都可以剪掉。
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/9a8c7ba6b5ae413799c6a33490bb1f8f~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/b1.png)
 
 所以呢，用了备忘录递归算法，递归树变成光秃秃的树干咯，如下：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/18a78d6b99a84ff8a1df834f3c085014~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/c1.png)
 
 带**备忘录**的递归算法，子问题个数=树节点数=n，解决一个子问题还是O(1),所以带**备忘录**的递归算法的时间复杂度是O(n)。接下来呢，我们用带**备忘录**的递归算法去撸代码，解决这个青蛙跳阶问题的超时问题咯~，代码如下：
 
@@ -174,7 +174,7 @@ kotlin
 
 去leetcode提交一下，如图，稳了：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/e5f0fb1998c542dc8cb3826de800f443~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/d1.png)
 
 其实，还可以用动态规划解决这道题。
 
@@ -194,11 +194,11 @@ kotlin
 
 我们来看下自底向上的解法，从f(1)往f(10）方向，想想是不是直接一个for循环就可以解决啦，如下：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/8e21b14d47e64afa958506d49972827b~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/e1.png)
 
 带备忘录的递归解法，空间复杂度是O(n)，但是呢，仔细观察上图，可以发现，f（n）只依赖前面两个数，所以只需要两个变量a和b来存储，就可以满足需求了，因此空间复杂度是O(1)就可以啦
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/8b202b2364eb4aae9122cfa8474c045e~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/f1.png)
 
 动态规划实现代码如下：
 
@@ -250,7 +250,7 @@ ini
 - 当台阶是4级时，想跳到第3级台阶，要么是先跳到第3级，然后再跳1级台阶上去，要么是先跳到第 2级，然后一次迈 2 级台阶上去。所以f(4) = f(3) + f(2) =5
 - 当台阶是5级时......
 
-![自底向上的动态规划](img/2024-03-04-%E5%A5%BD%E6%96%87/29cd6d0d31514336baf9905f8084a624~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![自底向上的动态规划](img/2024-03-04-%E5%A5%BD%E6%96%87/g1.png)
 
 #### 2. 确定边界
 
@@ -266,7 +266,7 @@ n>=3时，已经呈现出规律 f(n) = f(n-1) + f(n-2) ，因此，f(n-1)和f(n-
 
 通过前面3步，穷举分析，确定边界，最优子结构，我们就可以得出状态转移方程啦：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/4fb4cd7257ff4fcbac9d87be96f353bd~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/a2.png)
 
 #### 5. 代码实现
 
@@ -346,11 +346,11 @@ css
 
 是不是感觉成功了一半呢？但是**如何把nums[i]结尾的递增子序列也转化为对应的子问题**呢？要是nums[i]结尾的递增子序列也跟nums[i-1]的最长递增子序列有关就好了。又或者nums[i]结尾的最长递增子序列，跟前面子问题num[j]（0=<j<i）结尾的最长递增子序列有关就好了，带着这个想法，我们又回头看看穷举的过程：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/41d5ff02fc4b4da9bb6de5d3a1ba7c25~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/b2.png)
 
 nums[i]的最长递增子序列，不就是**从以数组num[i]每个元素结尾的最长子序列集合，取元素最多（也就是长度最长）那个嘛**，所以原问题，我们转化成求出以数组nums每个元素结尾的最长子序列集合，再取**最大值**嘛。哈哈，想到这，我们就可以**用dp[i]表示以num[i]这个数结尾的最长递增子序列的长度**啦，然后再来看看其中的规律：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/79287ac2dd7d47f9beb811d0c3630409~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/c2.png)
 
 其实，**nums[i]结尾的自增子序列，只要找到比nums[i]小的子序列，加上nums[i]** 就可以啦。显然，可能形成多种新的子序列，我们选最长那个，就是dp[i]的值啦
 
@@ -381,7 +381,7 @@ scss
 
 通过前面分析，我们就可以得出状态转移方程啦：
 
-![img](img/2024-03-04-%E5%A5%BD%E6%96%87/457ce281c1ec4e91b8c868aa593e41b6~tplv-k3u1fbpfcp-zoom-in-crop-mark:1512:0:0:0.awebp)
+![img](img/2024-03-04-%E5%A5%BD%E6%96%87/d2.png)
 
 所以数组num[i]的最长递增子序列就是：
 

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