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docs: update code blocks
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lcof/面试题03. 数组中重复的数字/README.md

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@@ -165,7 +165,7 @@ function swap(nums: number[], i: number, j: number): void {
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn find_repeat_number(mut nums: Vec<i32>) -> i32 {
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for i in 0..nums.len() {

lcof2/剑指 Offer II 035. 最小时间差/README.md

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<!-- 这里可写通用的实现逻辑 -->
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我们注意到,时间点最多只有 `24 * 60` 个,因此,当 timePoints 长度超过 `24 * 60`,说明有重复的时间点,提前返回 0。
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接下来:
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首先,遍历时间列表,将其转换为“分钟制”列表 `mins`,比如,对于时间点 `13:14`,将其转换为 `13 * 60 + 14`
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接着将“分钟制”列表按升序排列,然后将此列表的最小时间 `mins[0]` 加上 `24 * 60` 追加至列表尾部,用于处理最大值、最小值的差值这种特殊情况。

solution/0000-0099/0026.Remove Duplicates from Sorted Array/README.md

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@@ -65,8 +65,8 @@ for (int i = 0; i < len; i++) {
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原问题要求最多相同的数字最多出现 1 次,我们可以扩展至相同的数字最多保留 k 个。
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- 由于相同的数字最多保留 k 个,那么原数组的前 k 个元素我们可以直接保留;
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- 对于后面的数字,能够保留的前提是:当前数字 num 与前面已保留的数字的倒数第 k 个元素比较,不同则保留,相同则跳过。
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- 由于相同的数字最多保留 k 个,那么原数组的前 k 个元素我们可以直接保留;
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- 对于后面的数字,能够保留的前提是:当前数字 num 与前面已保留的数字的倒数第 k 个元素比较,不同则保留,相同则跳过。
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相似题目:[80. 删除有序数组中的重复项 II](/solution/0000-0099/0080.Remove%20Duplicates%20from%20Sorted%20Array%20II/README.md)
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn remove_duplicates(nums: &mut Vec<i32>) -> i32 {
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let mut len = 0;

solution/0000-0099/0026.Remove Duplicates from Sorted Array/README_EN.md

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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn remove_duplicates(nums: &mut Vec<i32>) -> i32 {
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let mut len = 0;

solution/0000-0099/0027.Remove Element/README.md

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@@ -153,7 +153,7 @@ func removeElement(nums []int, val int) int {
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn remove_element(nums: &mut Vec<i32>, val: i32) -> i32 {
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let mut len = 0;

solution/0000-0099/0027.Remove Element/README_EN.md

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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn remove_element(nums: &mut Vec<i32>, val: i32) -> i32 {
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let mut len = 0;

solution/0000-0099/0080.Remove Duplicates from Sorted Array II/README.md

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Original file line numberDiff line numberDiff line change
@@ -65,8 +65,8 @@ for (int i = 0; i < len; i++) {
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原问题要求最多相同的数字最多出现 2 次,我们可以扩展至相同的数字最多保留 k 个。
6767

68-
- 由于相同的数字最多保留 k 个,那么原数组的前 k 个元素我们可以直接保留;
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- 对于后面的数字,能够保留的前提是:当前数字 num 与前面已保留的数字的倒数第 k 个元素比较,不同则保留,相同则跳过。
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- 由于相同的数字最多保留 k 个,那么原数组的前 k 个元素我们可以直接保留;
69+
- 对于后面的数字,能够保留的前提是:当前数字 num 与前面已保留的数字的倒数第 k 个元素比较,不同则保留,相同则跳过。
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相似题目:[26. 删除有序数组中的重复项](/solution/0000-0099/0026.Remove%20Duplicates%20from%20Sorted%20Array/README.md)
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@@ -173,7 +173,7 @@ func removeDuplicates(nums []int) int {
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn remove_duplicates(nums: &mut Vec<i32>) -> i32 {
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let mut len = 0;

solution/0000-0099/0080.Remove Duplicates from Sorted Array II/README_EN.md

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@@ -154,7 +154,7 @@ func removeDuplicates(nums []int) int {
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn remove_duplicates(nums: &mut Vec<i32>) -> i32 {
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let mut len = 0;

solution/0100-0199/0169.Majority Element/README.md

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@@ -167,7 +167,7 @@ func majorityElement(nums []int) int {
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn majority_element(nums: Vec<i32>) -> i32 {
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let mut major = 0;

solution/0100-0199/0169.Majority Element/README_EN.md

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@@ -154,7 +154,7 @@ func majorityElement(nums []int) int {
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### **Rust**
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```rs
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```rust
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impl Solution {
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pub fn majority_element(nums: Vec<i32>) -> i32 {
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let mut major = 0;

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