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time complexity using go
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msk397 committed Dec 13, 2022
1 parent 58ca52d commit d4ad75e
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package time_complexity

/* 常数阶 */
func constant(n int) int {
count := 0
size := 100000
for i := 0; i < size; i++ {
count++
}
return count
}

/* 线性阶 */
func linear(n int) int {
count := 0
for i := 0; i < n; i++ {
count++
}
return count
}

/* 线性阶(遍历数组) */
func arrayTraversal(nums []int) int {
count := 0
// 循环次数与数组长度成正比
for range nums {
count++
}
return count
}

/* 平方阶 */
func quadratic(n int) int {
count := 0
// 循环次数与数组长度成平方关系
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
count++
}
}
return count
}

/* 平方阶(冒泡排序) */
func bubbleSort(nums []int) int {
count := 0 // 计数器
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
for i := len(nums) - 1; i > 0; i-- {
// 内循环:冒泡操作
for j := 0; j < i; j++ {
if nums[j] > nums[j+1] {
// 交换 nums[j] 与 nums[j + 1]
tmp := nums[j]
nums[j] = nums[j+1]
nums[j+1] = tmp
count += 3 // 元素交换包含 3 个单元操作
}
}
}
return count
}

/* 指数阶(循环实现)*/
func exponential(n int) int {
count, base := 0, 1
// cell 每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
for i := 0; i < n; i++ {
for j := 0; j < base; j++ {
count++
}
base *= 2
}
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
return count
}

/* 指数阶(递归实现)*/
func expRecur(n int) int {
if n == 1 {
return 1
}
return expRecur(n-1) + expRecur(n-1) + 1
}

/* 对数阶(循环实现)*/
func logarithmic(n float64) int {
count := 0
for n > 1 {
n = n / 2
count++
}
return count
}

/* 对数阶(递归实现)*/
func logRecur(n float64) int {
if n <= 1 {
return 0
}
return logRecur(n/2) + 1
}

/* 线性对数阶 */
func linearLogRecur(n float64) int {
if n <= 1 {
return 1
}
count := linearLogRecur(n/2) +
linearLogRecur(n/2)
for i := 0.0; i < n; i++ {
count++
}
return count
}

/* 阶乘阶(递归实现) */
func factorialRecur(n int) int {
if n == 0 {
return 1
}
count := 0
// 从 1 个分裂出 n 个
for i := 0; i < n; i++ {
count += factorialRecur(n - 1)
}
return count
}
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package time_complexity

import (
"fmt"
"testing"
)

func TestTimeComplexity(t *testing.T) {
n := 8
fmt.Println("输入数据大小 n =", n)

count := constant(n)
fmt.Println("常数阶的计算操作数量 =", count)

count = linear(n)
fmt.Println("线性阶的计算操作数量 =", count)
count = arrayTraversal(make([]int, n))
fmt.Println("线性阶(遍历数组)的计算操作数量 =", count)

count = quadratic(n)
fmt.Println("平方阶的计算操作数量 =", count)
nums := make([]int, n)
for i := 0; i < n; i++ {
nums[i] = n - i
}
count = bubbleSort(nums)
fmt.Println("平方阶(冒泡排序)的计算操作数量 =", count)

count = exponential(n)
fmt.Println("指数阶(循环实现)的计算操作数量 =", count)
count = expRecur(n)
fmt.Println("指数阶(递归实现)的计算操作数量 =", count)

count = logarithmic(float64(n))
fmt.Println("对数阶(循环实现)的计算操作数量 =", count)
count = logRecur(float64(n))
fmt.Println("对数阶(递归实现)的计算操作数量 =", count)

count = linearLogRecur(float64(n))
fmt.Println("线性对数阶(递归实现)的计算操作数量 =", count)

count = factorialRecur(n)
fmt.Println("阶乘阶(递归实现)的计算操作数量 =", count)
}
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