[Rust] No need to use mut iter (krahets#1356)

* No need to use mut iter

* Update iter

codes/rust/chapter_sorting/bucket_sort.rs

@@ -12,7 +12,7 @@ fn bucket_sort(nums: &mut [f64]) {
     let k = nums.len() / 2;
     let mut buckets = vec![vec![]; k];
     // 1. 将数组元素分配到各个桶中
-    for &mut num in &mut *nums {
+    for &num in nums.iter() {
         // 输入数据范围为 [0, 1)，使用 num * k 映射到索引范围 [0, k-1]
         let i = (num * k as f64) as usize;
         // 将 num 添加进桶 i
@@ -25,8 +25,8 @@ fn bucket_sort(nums: &mut [f64]) {
     }
     // 3. 遍历桶合并结果
     let mut i = 0;
-    for bucket in &mut buckets {
-        for &mut num in bucket {
+    for bucket in buckets.iter() {
+        for &num in bucket.iter() {
             nums[i] = num;
             i += 1;
         }

codes/rust/chapter_sorting/counting_sort.rs

@@ -10,11 +10,11 @@ include!("../include/include.rs");
 // 简单实现，无法用于排序对象
 fn counting_sort_naive(nums: &mut [i32]) {
     // 1. 统计数组最大元素 m
-    let m = *nums.into_iter().max().unwrap();
+    let m = *nums.iter().max().unwrap();
     // 2. 统计各数字的出现次数
     // counter[num] 代表 num 的出现次数
     let mut counter = vec![0; m as usize + 1];
-    for &num in &*nums {
+    for &num in nums.iter() {
         counter[num as usize] += 1;
     }
     // 3. 遍历 counter ，将各元素填入原数组 nums
@@ -31,16 +31,16 @@ fn counting_sort_naive(nums: &mut [i32]) {
 // 完整实现，可排序对象，并且是稳定排序
 fn counting_sort(nums: &mut [i32]) {
     // 1. 统计数组最大元素 m
-    let m = *nums.into_iter().max().unwrap();
+    let m = *nums.iter().max().unwrap() as usize;
     // 2. 统计各数字的出现次数
     // counter[num] 代表 num 的出现次数
-    let mut counter = vec![0; m as usize + 1];
-    for &num in &*nums {
+    let mut counter = vec![0; m + 1];
+    for &num in nums.iter() {
         counter[num as usize] += 1;
     }
     // 3. 求 counter 的前缀和，将“出现次数”转换为“尾索引”
     // 即 counter[num]-1 是 num 在 res 中最后一次出现的索引
-    for i in 0..m as usize {
+    for i in 0..m {
         counter[i + 1] += counter[i];
     }
     // 4. 倒序遍历 nums ，将各元素填入结果数组 res
@@ -53,9 +53,7 @@ fn counting_sort(nums: &mut [i32]) {
         counter[num as usize] -= 1; // 令前缀和自减 1 ，得到下次放置 num 的索引
     }
     // 使用结果数组 res 覆盖原数组 nums
-    for i in 0..n {
-        nums[i] = res[i];
-    }
+    nums.copy_from_slice(&res)
 }
 
 /* Driver Code */