hello-algo/zh-hant/codes/kotlin/chapter_computational_complexity/time_complexity.kt
Yudong Jin b2f0d4603d
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2024-04-11 20:18:19 +08:00

167 lines
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3.9 KiB
Kotlin

/**
* File: time_complexity.kt
* Created Time: 2024-01-25
* Author: curtishd (1023632660@qq.com)
*/
package chapter_computational_complexity.time_complexity
/* 常數階 */
fun constant(n: Int): Int {
var count = 0
val size = 100000
for (i in 0..<size)
count++
return count
}
/* 線性階 */
fun linear(n: Int): Int {
var count = 0
// 迴圈次數與陣列長度成正比
for (i in 0..<n)
count++
return count
}
/* 線性階(走訪陣列) */
fun arrayTraversal(nums: IntArray): Int {
var count = 0
// 迴圈次數與陣列長度成正比
for (num in nums) {
count++
}
return count
}
/* 平方階 */
fun quadratic(n: Int): Int {
var count = 0
// 迴圈次數與資料大小 n 成平方關係
for (i in 0..<n) {
for (j in 0..<n) {
count++
}
}
return count
}
/* 平方階(泡沫排序) */
fun bubbleSort(nums: IntArray): Int {
var count = 0 // 計數器
// 外迴圈:未排序區間為 [0, i]
for (i in nums.size - 1 downTo 1) {
// 內迴圈:將未排序區間 [0, i] 中的最大元素交換至該區間的最右端
for (j in 0..<i) {
if (nums[j] > nums[j + 1]) {
// 交換 nums[j] 與 nums[j + 1]
nums[j] = nums[j + 1].also { nums[j + 1] = nums[j] }
count += 3 // 元素交換包含 3 個單元操作
}
}
}
return count
}
/* 指數階(迴圈實現) */
fun exponential(n: Int): Int {
var count = 0
// 細胞每輪一分為二,形成數列 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
}
/* 指數階(遞迴實現) */
fun expRecur(n: Int): Int {
if (n == 1) {
return 1
}
return expRecur(n - 1) + expRecur(n - 1) + 1
}
/* 對數階(迴圈實現) */
fun logarithmic(n: Int): Int {
var n1 = n
var count = 0
while (n1 > 1) {
n1 /= 2
count++
}
return count
}
/* 對數階(遞迴實現) */
fun logRecur(n: Int): Int {
if (n <= 1)
return 0
return logRecur(n / 2) + 1
}
/* 線性對數階 */
fun linearLogRecur(n: Int): Int {
if (n <= 1)
return 1
var count = linearLogRecur(n / 2) + linearLogRecur(n / 2)
for (i in 0..<n) {
count++
}
return count
}
/* 階乘階(遞迴實現) */
fun factorialRecur(n: Int): Int {
if (n == 0)
return 1
var count = 0
// 從 1 個分裂出 n 個
for (i in 0..<n) {
count += factorialRecur(n - 1)
}
return count
}
/* Driver Code */
fun main() {
// 可以修改 n 執行,體會一下各種複雜度的操作數量變化趨勢
val n = 8
println("輸入資料大小 n = $n")
var count = constant(n)
println("常數階的操作數量 = $count")
count = linear(n)
println("線性階的操作數量 = $count")
count = arrayTraversal(IntArray(n))
println("線性階(走訪陣列)的操作數量 = $count")
count = quadratic(n)
println("平方階的操作數量 = $count")
val nums = IntArray(n)
for (i in 0..<n)
nums[i] = n - i // [n,n-1,...,2,1]
count = bubbleSort(nums)
println("平方階(泡沫排序)的操作數量 = $count")
count = exponential(n)
println("指數階(迴圈實現)的操作數量 = $count")
count = expRecur(n)
println("指數階(遞迴實現)的操作數量 = $count")
count = logarithmic(n)
println("對數階(迴圈實現)的操作數量 = $count")
count = logRecur(n)
println("對數階(遞迴實現)的操作數量 = $count")
count = linearLogRecur(n)
println("線性對數階(遞迴實現)的操作數量 = $count")
count = factorialRecur(n)
println("階乘階(遞迴實現)的操作數量 = $count")
}