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lcci/16.10.Living People/README.md

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@@ -116,7 +116,6 @@ function maxAliveYear(birth: number[], death: number[]): number {
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}
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return res;
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}
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```
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### **Rust**

lcci/16.10.Living People/README_EN.md

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}
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return res;
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}
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```
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### **Rust**

lcp/LCP 34. 二叉树染色/README.md

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@@ -61,7 +61,7 @@ class Solution:
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for j in range(k + 1):
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ans[0] = max(ans[0], l[i] + r[j])
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return ans
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return max(dfs(root))
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```
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solution/0200-0299/0241.Different Ways to Add Parentheses/README.md

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else:
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ans.append(a * b)
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return ans
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return dfs(expression)
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```
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solution/0200-0299/0241.Different Ways to Add Parentheses/README_EN.md

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else:
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ans.append(a * b)
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return ans
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return dfs(expression)
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```
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solution/0300-0399/0336.Palindrome Pairs/README.md

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<strong>输出:</strong>[[0,1],[1,0]]
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</pre>
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<p><strong>提示:</strong></p>
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<ul>

solution/0500-0599/0540.Single Element in a Sorted Array/README.md

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@@ -58,8 +58,8 @@ for (int i = 0; i < n - 1; i += 2) {
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return nums[n - 1];
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```
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偶数下标:当 `nums[i] != nums[i + 1] && i % 2 == 0` 成立,结果便是 `nums[i]`
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奇数下标:当 `nums[i] != nums[i - 1] && i % 2 == 1` 成立,结果便是 `nums[i - 1]`
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偶数下标:当 `nums[i] != nums[i + 1] && i % 2 == 0` 成立,结果便是 `nums[i]`
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奇数下标:当 `nums[i] != nums[i - 1] && i % 2 == 1` 成立,结果便是 `nums[i - 1]`
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于是二分模板就有了:
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solution/0500-0599/0554.Brick Wall/README.md

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<strong>输出:</strong>3
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</pre>
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<p><strong>提示:</strong></p>
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<ul>

solution/0700-0799/0745.Prefix and Suffix Search/README.md

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@@ -31,8 +31,6 @@ WordFilter wordFilter = new WordFilter(["apple"]);
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wordFilter.f("a", "e"); // 返回 0 ,因为下标为 0 的单词的 prefix = "a" 且 suffix = 'e" 。
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</pre>
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<p><strong>提示:</strong></p>
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solution/0700-0799/0775.Global and Local Inversions/README.md

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<strong>解释:</strong>有 2 个全局倒置,和 1 个局部倒置。
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</pre>
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<p><strong>提示:</strong></p>
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<ul>

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