Merge branch 'krahets:main' into main
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4b598b0e6a
42
codes/ruby/chapter_divide_and_conquer/binary_search_recur.rb
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codes/ruby/chapter_divide_and_conquer/binary_search_recur.rb
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=begin
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File: binary_search_recur.rb
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Created Time: 2024-05-13
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 二分查找:问题 f(i, j) ###
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def dfs(nums, target, i, j)
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# 若区间为空,代表无目标元素,则返回 -1
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return -1 if i > j
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# 计算中点索引 m
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m = (i + j) / 2
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if nums[m] < target
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# 递归子问题 f(m+1, j)
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return dfs(nums, target, m + 1, j)
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elsif nums[m] > target
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# 递归子问题 f(i, m-1)
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return dfs(nums, target, i, m - 1)
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else
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# 找到目标元素,返回其索引
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return m
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end
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end
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### 二分查找 ###
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def binary_search(nums, target)
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n = nums.length
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# 求解问题 f(0, n-1)
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dfs(nums, target, 0, n - 1)
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end
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### Driver Code ###
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if __FILE__ == $0
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target = 6
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nums = [1, 3, 6, 8, 12, 15, 23, 26, 31, 35]
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# 二分查找(双闭区间)
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index = binary_search(nums, target)
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puts "目标元素 6 的索引 = #{index}"
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end
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46
codes/ruby/chapter_divide_and_conquer/build_tree.rb
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codes/ruby/chapter_divide_and_conquer/build_tree.rb
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=begin
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File: build_tree.rb
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Created Time: 2024-05-13
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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require_relative '../utils/tree_node'
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require_relative '../utils/print_util'
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### 构建二叉树:分治 ###
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def dfs(preorder, inorder_map, i, l, r)
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# 子树区间为空时终止
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return if r - l < 0
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# 初始化根节点
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root = TreeNode.new(preorder[i])
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# 查询 m ,从而划分左右子树
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m = inorder_map[preorder[i]]
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# 子问题:构建左子树
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root.left = dfs(preorder, inorder_map, i + 1, l, m - 1)
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# 子问题:构建右子树
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root.right = dfs(preorder, inorder_map, i + 1 + m - l, m + 1, r)
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# 返回根节点
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root
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end
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### 构建二叉树 ###
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def build_tree(preorder, inorder)
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# 初始化哈希表,存储 inorder 元素到索引的映射
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inorder_map = {}
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inorder.each_with_index { |val, i| inorder_map[val] = i }
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dfs(preorder, inorder_map, 0, 0, inorder.length - 1)
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end
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### Driver Code ###
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if __FILE__ == $0
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preorder = [3, 9, 2, 1, 7]
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inorder = [9, 3, 1, 2, 7]
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puts "前序遍历 = #{preorder}"
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puts "中序遍历 = #{inorder}"
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root = build_tree(preorder, inorder)
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puts "构建的二叉树为:"
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print_tree(root)
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end
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55
codes/ruby/chapter_divide_and_conquer/hanota.rb
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codes/ruby/chapter_divide_and_conquer/hanota.rb
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=begin
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File: hanota.rb
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Created Time: 2024-05-13
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Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
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=end
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### 移动一个圆盘 ###
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def move(src, tar)
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# 从 src 顶部拿出一个圆盘
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pan = src.pop
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# 将圆盘放入 tar 顶部
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tar << pan
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end
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### 求解汉诺塔问题 f(i) ###
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def dfs(i, src, buf, tar)
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# 若 src 只剩下一个圆盘,则直接将其移到 tar
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if i == 1
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move(src, tar)
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return
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end
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# 子问题 f(i-1) :将 src 顶部 i-1 个圆盘借助 tar 移到 buf
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dfs(i - 1, src, tar, buf)
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# 子问题 f(1) :将 src 剩余一个圆盘移到 tar
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move(src, tar)
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# 子问题 f(i-1) :将 buf 顶部 i-1 个圆盘借助 src 移到 tar
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dfs(i - 1, buf, src, tar)
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end
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### 求解汉诺塔问题 ###
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def solve_hanota(_A, _B, _C)
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n = _A.length
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# 将 A 顶部 n 个圆盘借助 B 移到 C
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dfs(n, _A, _B, _C)
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end
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### Driver Code ###
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if __FILE__ == $0
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# 列表尾部是柱子顶部
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A = [5, 4, 3, 2, 1]
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B = []
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C = []
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puts "初始状态下:"
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puts "A = #{A}"
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puts "B = #{B}"
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puts "C = #{C}"
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solve_hanota(A, B, C)
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puts "圆盘移动完成后:"
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puts "A = #{A}"
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puts "B = #{B}"
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puts "C = #{C}"
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end
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@ -9,7 +9,7 @@
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/* 移动一个圆盘 */
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fn move_pan(src: &mut Vec<i32>, tar: &mut Vec<i32>) {
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// 从 src 顶部拿出一个圆盘
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let pan = src.remove(src.len() - 1);
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let pan = src.pop().unwrap();
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// 将圆盘放入 tar 顶部
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tar.push(pan);
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}
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@ -14,9 +14,7 @@ fn bubble_sort(nums: &mut [i32]) {
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for j in 0..i {
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if nums[j] > nums[j + 1] {
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// 交换 nums[j] 与 nums[j + 1]
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let tmp = nums[j];
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nums[j] = nums[j + 1];
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nums[j + 1] = tmp;
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nums.swap(j, j + 1);
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}
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}
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}
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@ -31,9 +29,7 @@ fn bubble_sort_with_flag(nums: &mut [i32]) {
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for j in 0..i {
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if nums[j] > nums[j + 1] {
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// 交换 nums[j] 与 nums[j + 1]
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let tmp = nums[j];
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nums[j] = nums[j + 1];
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nums[j + 1] = tmp;
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nums.swap(j, j + 1);
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flag = true; // 记录交换元素
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}
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}
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@ -12,7 +12,7 @@ fn bucket_sort(nums: &mut [f64]) {
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let k = nums.len() / 2;
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let mut buckets = vec![vec![]; k];
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// 1. 将数组元素分配到各个桶中
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for &mut num in &mut *nums {
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for &num in nums.iter() {
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// 输入数据范围为 [0, 1),使用 num * k 映射到索引范围 [0, k-1]
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let i = (num * k as f64) as usize;
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// 将 num 添加进桶 i
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@ -25,8 +25,8 @@ fn bucket_sort(nums: &mut [f64]) {
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}
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// 3. 遍历桶合并结果
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let mut i = 0;
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for bucket in &mut buckets {
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for &mut num in bucket {
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for bucket in buckets.iter() {
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for &num in bucket.iter() {
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nums[i] = num;
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i += 1;
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}
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// 简单实现,无法用于排序对象
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fn counting_sort_naive(nums: &mut [i32]) {
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// 1. 统计数组最大元素 m
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let m = *nums.into_iter().max().unwrap();
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let m = *nums.iter().max().unwrap();
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// 2. 统计各数字的出现次数
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// counter[num] 代表 num 的出现次数
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let mut counter = vec![0; m as usize + 1];
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for &num in &*nums {
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for &num in nums.iter() {
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counter[num as usize] += 1;
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}
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// 3. 遍历 counter ,将各元素填入原数组 nums
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@ -31,16 +31,16 @@ fn counting_sort_naive(nums: &mut [i32]) {
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// 完整实现,可排序对象,并且是稳定排序
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fn counting_sort(nums: &mut [i32]) {
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// 1. 统计数组最大元素 m
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let m = *nums.into_iter().max().unwrap();
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let m = *nums.iter().max().unwrap() as usize;
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// 2. 统计各数字的出现次数
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// counter[num] 代表 num 的出现次数
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let mut counter = vec![0; m as usize + 1];
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for &num in &*nums {
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let mut counter = vec![0; m + 1];
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for &num in nums.iter() {
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counter[num as usize] += 1;
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}
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// 3. 求 counter 的前缀和,将“出现次数”转换为“尾索引”
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// 即 counter[num]-1 是 num 在 res 中最后一次出现的索引
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for i in 0..m as usize {
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for i in 0..m {
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counter[i + 1] += counter[i];
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}
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// 4. 倒序遍历 nums ,将各元素填入结果数组 res
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@ -53,9 +53,7 @@ fn counting_sort(nums: &mut [i32]) {
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counter[num as usize] -= 1; // 令前缀和自减 1 ,得到下次放置 num 的索引
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}
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// 使用结果数组 res 覆盖原数组 nums
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for i in 0..n {
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nums[i] = res[i];
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}
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nums.copy_from_slice(&res)
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}
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/* Driver Code */
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@ -24,9 +24,7 @@ fn sift_down(nums: &mut [i32], n: usize, mut i: usize) {
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break;
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}
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// 交换两节点
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let temp = nums[i];
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nums[i] = nums[ma];
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nums[ma] = temp;
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nums.swap(i, ma);
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// 循环向下堆化
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i = ma;
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}
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@ -35,15 +33,13 @@ fn sift_down(nums: &mut [i32], n: usize, mut i: usize) {
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/* 堆排序 */
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fn heap_sort(nums: &mut [i32]) {
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// 建堆操作:堆化除叶节点以外的其他所有节点
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for i in (0..=nums.len() / 2 - 1).rev() {
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for i in (0..nums.len() / 2).rev() {
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sift_down(nums, nums.len(), i);
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}
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// 从堆中提取最大元素,循环 n-1 轮
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for i in (1..=nums.len() - 1).rev() {
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for i in (1..nums.len()).rev() {
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// 交换根节点与最右叶节点(交换首元素与尾元素)
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let tmp = nums[0];
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nums[0] = nums[i];
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nums[i] = tmp;
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nums.swap(0, i);
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// 以根节点为起点,从顶至底进行堆化
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sift_down(nums, i, 0);
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}
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