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Update 1143. 最长公共子序列.md
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Solutions/1143. 最长公共子序列.md

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return dp[size1][size2]
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```
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### 思路 1 :复杂度分析
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### 思路 1:复杂度分析
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- **时间复杂度**:$O(n^2)$。两重循环遍历的时间复杂度是 $O(n^2)$,所以总的时间复杂度为 $O(n^2)$。
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- **空间复杂度**:$O(n^2)$。用到了二维数组保存状态,所以总体空间复杂度为 $O(n^2)$。

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