hello-algo/codes/kotlin/chapter_computational_complexity/time_complexity.kt

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Kotlin

/**
* 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
// 循环次数与数据大小 n 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")
}