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Add Kotlin code for computational complexity (#1090)
* feat(kotlin):new kotlin support files
* fix(kotlin):
reviewed the formatting, comments and so on.
* fix(kotlin): fix the indentation and format
* feat(kotlin): Add kotlin code for the backtraking chapter.
* fix(kotlin): fix incorrect output of preorder_traversal_iii_template.kt file
* fix(kotlin): simplify kotlin codes
* fix(kotlin): modify n_queens.kt for consistency.
* feat(kotlin): add kotlin code for computational complexity.
* fix(kotlin): remove iteration folder.
* fix(kotlin): remove n_queens.kt file out of folder.
* fix(kotlin): remove some folders.
* style(kotlin): modified two chapters.
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/**
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* File: space_complexity.kt
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* Created Time: 2024-01-25
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* Author: curtishd (1023632660@qq.com)
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*/
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package chapter_computational_complexity.space_complexity
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import utils.ListNode
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import utils.TreeNode
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import utils.printTree
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/* 函数 */
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fun function(): Int {
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// 执行某些操作
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return 0
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}
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/* 常数阶 */
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fun constant(n: Int) {
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// 常量、变量、对象占用 O(1) 空间
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val a = 0
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var b = 0
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val nums = Array(10000) { 0 }
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val node = ListNode(0)
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// 循环中的变量占用 O(1) 空间
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for (i in 0..<n) {
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val c = 0
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}
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// 循环中的函数占用 O(1) 空间
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for (i in 0..<n) {
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function()
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}
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}
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/* 线性阶 */
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fun linear(n: Int) {
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// 长度为 n 的数组占用 O(n) 空间
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val nums = Array(n) { 0 }
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// 长度为 n 的列表占用 O(n) 空间
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val nodes = mutableListOf<ListNode>()
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for (i in 0..<n) {
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nodes.add(ListNode(i))
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}
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// 长度为 n 的哈希表占用 O(n) 空间
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val map = mutableMapOf<Int, String>()
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for (i in 0..<n) {
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map[i] = i.toString()
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}
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}
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/* 线性阶(递归实现) */
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fun linearRecur(n: Int) {
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println("递归 n = $n")
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if (n == 1)
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return
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linearRecur(n - 1)
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}
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/* 平方阶 */
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fun quadratic(n: Int) {
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// 矩阵占用 O(n^2) 空间
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val numMatrix: Array<Array<Int>?> = arrayOfNulls(n)
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// 二维列表占用 O(n^2) 空间
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val numList: MutableList<MutableList<Int>> = arrayListOf()
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for (i in 0..<n) {
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val tmp = mutableListOf<Int>()
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for (j in 0..<n) {
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tmp.add(0)
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}
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numList.add(tmp)
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}
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}
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/* 平方阶(递归实现) */
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tailrec fun quadraticRecur(n: Int): Int {
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if (n <= 0)
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return 0
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// 数组 nums 长度为 n, n-1, ..., 2, 1
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val nums = Array(n) { 0 }
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println("递归 n = $n 中的 nums 长度 = ${nums.size}")
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return quadraticRecur(n - 1)
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}
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/* 指数阶(建立满二叉树) */
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fun buildTree(n: Int): TreeNode? {
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if (n == 0)
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return null
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val root = TreeNode(0)
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root.left = buildTree(n - 1)
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root.right = buildTree(n - 1)
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return root
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}
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/* Driver Code */
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fun main() {
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val n = 5
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// 常数阶
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constant(n)
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// 线性阶
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linear(n)
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linearRecur(n)
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// 平方阶
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quadratic(n)
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quadraticRecur(n)
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// 指数阶
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val root: TreeNode? = buildTree(n)
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printTree(root)
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}
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