2024-02-27 17:04:57 +08:00
|
|
|
/**
|
|
|
|
* 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
|
2024-04-07 01:31:58 +08:00
|
|
|
val size = 100000
|
2024-02-27 17:04:57 +08:00
|
|
|
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
|
2024-03-19 02:28:16 +08:00
|
|
|
// 循环次数与数据大小 n 成平方关系
|
2024-02-27 17:04:57 +08:00
|
|
|
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 {
|
2024-04-07 01:31:58 +08:00
|
|
|
var count = 0 // 计数器
|
2024-02-27 17:04:57 +08:00
|
|
|
// 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) */
|
2024-03-23 02:17:48 +08:00
|
|
|
fun logarithmic(n: Int): Int {
|
2024-02-27 17:04:57 +08:00
|
|
|
var n1 = n
|
|
|
|
var count = 0
|
|
|
|
while (n1 > 1) {
|
|
|
|
n1 /= 2
|
|
|
|
count++
|
|
|
|
}
|
|
|
|
return count
|
|
|
|
}
|
|
|
|
|
|
|
|
/* 对数阶(递归实现) */
|
2024-03-23 02:17:48 +08:00
|
|
|
fun logRecur(n: Int): Int {
|
2024-02-27 17:04:57 +08:00
|
|
|
if (n <= 1)
|
|
|
|
return 0
|
|
|
|
return logRecur(n / 2) + 1
|
|
|
|
}
|
|
|
|
|
|
|
|
/* 线性对数阶 */
|
2024-03-23 02:17:48 +08:00
|
|
|
fun linearLogRecur(n: Int): Int {
|
2024-02-27 17:04:57 +08:00
|
|
|
if (n <= 1)
|
|
|
|
return 1
|
|
|
|
var count = linearLogRecur(n / 2) + linearLogRecur(n / 2)
|
2024-04-07 01:31:58 +08:00
|
|
|
for (i in 0..<n) {
|
2024-02-27 17:04:57 +08:00
|
|
|
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")
|
|
|
|
|
2024-04-07 01:31:58 +08:00
|
|
|
var count = constant(n)
|
2024-02-27 17:04:57 +08:00
|
|
|
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)
|
2024-04-07 01:31:58 +08:00
|
|
|
for (i in 0..<n)
|
|
|
|
nums[i] = n - i // [n,n-1,...,2,1]
|
2024-02-27 17:04:57 +08:00
|
|
|
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")
|
|
|
|
|
2024-03-23 02:17:48 +08:00
|
|
|
count = logarithmic(n)
|
2024-02-27 17:04:57 +08:00
|
|
|
println("对数阶(循环实现)的操作数量 = $count")
|
2024-03-23 02:17:48 +08:00
|
|
|
count = logRecur(n)
|
2024-02-27 17:04:57 +08:00
|
|
|
println("对数阶(递归实现)的操作数量 = $count")
|
|
|
|
|
2024-03-23 02:17:48 +08:00
|
|
|
count = linearLogRecur(n)
|
2024-02-27 17:04:57 +08:00
|
|
|
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")
|
|
|
|
}
|