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2019-03-18 10:34:46 +08:00
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docs/notes/Leetcode 题解 - 动态规划.md
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@ -567,7 +567,7 @@ public int wiggleMaxLength(int[] nums) {
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?dp[i][j]=\left\{\begin{array}{rcl}dp[i-1][j-1]&&{S1_i==S2_j}\\max(dp[i-1][j],dp[i][j-1])&&{S1_i<>S2_j}\end{array}\right."/></div> <br>-->
<div align="center"> <img src="pics/6c0cf1e8-b03f-4eff-9b1a-ab262e0c7866.png"/> </div><br>
<div align="center"> <img src="pics/ecd89a22-c075-4716-8423-e0ba89230e9a.jpg"/> </div><br>
对于长度为 N 的序列 S<sub>1</sub> 和长度为 M 的序列 S<sub>2</sub>，dp[N][M] 就是序列 S<sub>1</sub> 和序列 S<sub>2</sub> 的最长公共子序列长度。
@ -1223,4 +1223,5 @@ public int minSteps(int n) {
</br><div align="center"> <img src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7%20%E6%B5%B7%E6%8A%A5.png" width="500px"> </div></br>
</br><div align="center">公众号 CyC2018，专注于核心基础知识分享、求职指导、技术成长。在公众号后台回复 ziliao 可领取复习大纲，帮你理清复习重点。</div></br>
<div align="center"><img width="180px" src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg"></img></div>