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.

codes/kotlin/chapter_computational_complexity/time_complexity.kt

@@ -0,0 +1,166 @@
+/**
+ * File: time_complexity.kt
+ * Created Time: 2024-01-25
+ * Author: curtishd (1023632660@qq.com)
+ */
+
+package chapter_computational_complexity.time_complexity
+
+/* 常数阶 */
+fun constant(n: Int): Int {
+    var count = 0
+    val size = 10_0000
+    for (i in 0..<size)
+        count++
+    return count
+}
+
+/* 线性阶 */
+fun linear(n: Int): Int {
+    var count = 0
+    // 循环次数与数组长度成正比
+    for (i in 0..<n)
+        count++
+    return count
+}
+
+/* 线性阶（遍历数组） */
+fun arrayTraversal(nums: IntArray): Int {
+    var count = 0
+    // 循环次数与数组长度成正比
+    for (num in nums) {
+        count++
+    }
+    return count
+}
+
+/* 平方阶 */
+fun quadratic(n: Int): Int {
+    var count = 0
+    // 循环次数与数组长度成平方关系
+    for (i in 0..<n) {
+        for (j in 0..<n) {
+            count++
+        }
+    }
+    return count
+}
+
+/* 平方阶（冒泡排序） */
+fun bubbleSort(nums: IntArray): Int {
+    var count = 0
+    // 外循环：未排序区间为 [0, i]
+    for (i in nums.size - 1 downTo 1) {
+        // 内循环：将未排序区间 [0, i] 中的最大元素交换至该区间的最右端
+        for (j in 0..<i) {
+            if (nums[j] > nums[j + 1]) {
+                // 交换 nums[j] 与 nums[j + 1]
+                nums[j] = nums[j + 1].also { nums[j + 1] = nums[j] }
+                count += 3 // 元素交换包含 3 个单元操作
+            }
+        }
+    }
+    return count
+}
+
+/* 指数阶（循环实现） */
+fun exponential(n: Int): Int {
+    var count = 0
+    // 细胞每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
+    var base = 1
+    for (i in 0..<n) {
+        for (j in 0..<base) {
+            count++
+        }
+        base *= 2
+    }
+    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
+    return count
+}
+
+/* 指数阶（递归实现） */
+fun expRecur(n: Int): Int {
+    if (n == 1) {
+        return 1
+    }
+    return expRecur(n - 1) + expRecur(n - 1) + 1
+}
+
+/* 对数阶（循环实现） */
+fun logarithmic(n: Float): Int {
+    var n1 = n
+    var count = 0
+    while (n1 > 1) {
+        n1 /= 2
+        count++
+    }
+    return count
+}
+
+/* 对数阶（递归实现） */
+fun logRecur(n: Float): Int {
+    if (n <= 1)
+        return 0
+    return logRecur(n / 2) + 1
+}
+
+/* 线性对数阶 */
+fun linearLogRecur(n: Float): Int {
+    if (n <= 1)
+        return 1
+    var count = linearLogRecur(n / 2) + linearLogRecur(n / 2)
+    for (i in 0..<n.toInt()) {
+        count++
+    }
+    return count
+}
+
+/* 阶乘阶（递归实现） */
+fun factorialRecur(n: Int): Int {
+    if (n == 0)
+        return 1
+    var count = 0
+    // 从 1 个分裂出 n 个
+    for (i in 0..<n) {
+        count += factorialRecur(n - 1)
+    }
+    return count
+}
+
+/* Driver Code */
+fun main() {
+    // 可以修改 n 运行，体会一下各种复杂度的操作数量变化趋势
+    val n = 8
+    println("输入数据大小 n = $n")
+
+    var count: Int = constant(n)
+    println("常数阶的操作数量 = $count")
+
+    count = linear(n)
+    println("线性阶的操作数量 = $count")
+    count = arrayTraversal(IntArray(n))
+    println("线性阶（遍历数组）的操作数量 = $count")
+
+    count = quadratic(n)
+    println("平方阶的操作数量 = $count")
+    val nums = IntArray(n)
+    for (i in 0..<n) nums[i] = n - i // [n,n-1,...,2,1]
+    count = bubbleSort(nums)
+    println("平方阶（冒泡排序）的操作数量 = $count")
+
+    count = exponential(n)
+    println("指数阶（循环实现）的操作数量 = $count")
+    count = expRecur(n)
+    println("指数阶（递归实现）的操作数量 = $count")
+
+    count = logarithmic(n.toFloat())
+    println("对数阶（循环实现）的操作数量 = $count")
+    count = logRecur(n.toFloat())
+    println("对数阶（递归实现）的操作数量 = $count")
+
+    count = linearLogRecur(n.toFloat())
+    println("线性对数阶（递归实现）的操作数量 = $count")
+
+    count = factorialRecur(n)
+    println("阶乘阶（递归实现）的操作数量 = $count")
+}