mirror of
				https://github.com/krahets/hello-algo.git
				synced 2025-11-04 22:28:40 +08:00 
			
		
		
		
	* Use int instead of float for the example code of log time complexity * Bug fixes * Bug fixes
		
			
				
	
	
		
			180 lines
		
	
	
		
			4.6 KiB
		
	
	
	
		
			Zig
		
	
	
	
	
	
			
		
		
	
	
			180 lines
		
	
	
		
			4.6 KiB
		
	
	
	
		
			Zig
		
	
	
	
	
	
// File: time_complexity.zig
 | 
						|
// Created Time: 2022-12-28
 | 
						|
// Author: codingonion (coderonion@gmail.com)
 | 
						|
 | 
						|
const std = @import("std");
 | 
						|
 | 
						|
// 常数阶
 | 
						|
fn constant(n: i32) i32 {
 | 
						|
    _ = n;
 | 
						|
    var count: i32 = 0;
 | 
						|
    const size: i32 = 100_000;
 | 
						|
    var i: i32 = 0;
 | 
						|
    while(i<size) : (i += 1) {
 | 
						|
        count += 1;
 | 
						|
    }
 | 
						|
    return count;
 | 
						|
}
 | 
						|
 | 
						|
// 线性阶
 | 
						|
fn linear(n: i32) i32 {
 | 
						|
    var count: i32 = 0;
 | 
						|
    var i: i32 = 0;
 | 
						|
    while (i < n) : (i += 1) {
 | 
						|
        count += 1;
 | 
						|
    }
 | 
						|
    return count;
 | 
						|
}
 | 
						|
 | 
						|
// 线性阶(遍历数组)
 | 
						|
fn arrayTraversal(nums: []i32) i32 {
 | 
						|
    var count: i32 = 0;
 | 
						|
    // 循环次数与数组长度成正比
 | 
						|
    for (nums) |_| {
 | 
						|
        count += 1;
 | 
						|
    }
 | 
						|
    return count;
 | 
						|
}
 | 
						|
 | 
						|
// 平方阶
 | 
						|
fn quadratic(n: i32) i32 {
 | 
						|
    var count: i32 = 0;
 | 
						|
    var i: i32 = 0;
 | 
						|
    // 循环次数与数据大小 n 成平方关系
 | 
						|
    while (i < n) : (i += 1) {
 | 
						|
        var j: i32 = 0;
 | 
						|
        while (j < n) : (j += 1) {
 | 
						|
            count += 1;
 | 
						|
        }
 | 
						|
    }
 | 
						|
    return count;
 | 
						|
}
 | 
						|
 | 
						|
// 平方阶(冒泡排序)
 | 
						|
fn bubbleSort(nums: []i32) i32 {
 | 
						|
    var count: i32 = 0;  // 计数器 
 | 
						|
    // 外循环:未排序区间为 [0, i]
 | 
						|
    var i: i32 = @as(i32, @intCast(nums.len)) - 1;
 | 
						|
    while (i > 0) : (i -= 1) {
 | 
						|
        var j: usize = 0;
 | 
						|
        // 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
 | 
						|
        while (j < i) : (j += 1) {
 | 
						|
            if (nums[j] > nums[j + 1]) {
 | 
						|
                // 交换 nums[j] 与 nums[j + 1]
 | 
						|
                var 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)
 | 
						|
fn exponential(n: i32) i32 {
 | 
						|
    var count: i32 = 0;
 | 
						|
    var bas: i32 = 1;
 | 
						|
    var i: i32 = 0;
 | 
						|
    // 细胞每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
 | 
						|
    while (i < n) : (i += 1) {
 | 
						|
        var j: i32 = 0;
 | 
						|
        while (j < bas) : (j += 1) {
 | 
						|
            count += 1;
 | 
						|
        }
 | 
						|
        bas *= 2;
 | 
						|
    }
 | 
						|
    // count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
 | 
						|
    return count;
 | 
						|
}
 | 
						|
 | 
						|
// 指数阶(递归实现)
 | 
						|
fn expRecur(n: i32) i32 {
 | 
						|
    if (n == 1) return 1;
 | 
						|
    return expRecur(n - 1) + expRecur(n - 1) + 1;
 | 
						|
}
 | 
						|
 | 
						|
// 对数阶(循环实现)
 | 
						|
fn logarithmic(n: i32) i32 {
 | 
						|
    var count: i32 = 0;
 | 
						|
    var n_var = n;
 | 
						|
    while (n_var > 1)
 | 
						|
    {
 | 
						|
        n_var = n_var / 2;
 | 
						|
        count +=1;
 | 
						|
    }
 | 
						|
    return count;
 | 
						|
}
 | 
						|
 | 
						|
// 对数阶(递归实现)
 | 
						|
fn logRecur(n: i32) i32 {
 | 
						|
    if (n <= 1) return 0;
 | 
						|
    return logRecur(n / 2) + 1;
 | 
						|
}
 | 
						|
 | 
						|
// 线性对数阶
 | 
						|
fn linearLogRecur(n: i32) i32 {
 | 
						|
    if (n <= 1) return 1;
 | 
						|
    var count: i32 = linearLogRecur(n / 2) + linearLogRecur(n / 2);
 | 
						|
    var i: i32 = 0;
 | 
						|
    while (i < n) : (i += 1) {
 | 
						|
        count += 1;
 | 
						|
    }
 | 
						|
    return count;
 | 
						|
}
 | 
						|
 | 
						|
// 阶乘阶(递归实现)
 | 
						|
fn factorialRecur(n: i32) i32 {
 | 
						|
    if (n == 0) return 1;
 | 
						|
    var count: i32 = 0;
 | 
						|
    var i: i32 = 0;
 | 
						|
    // 从 1 个分裂出 n 个
 | 
						|
    while (i < n) : (i += 1) {
 | 
						|
        count += factorialRecur(n - 1);
 | 
						|
    }
 | 
						|
    return count;
 | 
						|
}
 | 
						|
 | 
						|
// Driver Code
 | 
						|
pub fn main() !void {
 | 
						|
    // 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
 | 
						|
    const n: i32 = 8;
 | 
						|
    std.debug.print("输入数据大小 n = {}\n", .{n});
 | 
						|
 | 
						|
    var count = constant(n);
 | 
						|
    std.debug.print("常数阶的操作数量 = {}\n", .{count});
 | 
						|
 | 
						|
    count = linear(n);
 | 
						|
    std.debug.print("线性阶的操作数量 = {}\n", .{count});
 | 
						|
    var nums = [_]i32{0}**n;
 | 
						|
    count = arrayTraversal(&nums);
 | 
						|
    std.debug.print("线性阶(遍历数组)的操作数量 = {}\n", .{count});
 | 
						|
 | 
						|
    count = quadratic(n);
 | 
						|
    std.debug.print("平方阶的操作数量 = {}\n", .{count});
 | 
						|
    for (&nums, 0..) |*num, i| {
 | 
						|
        num.* = n - @as(i32, @intCast(i));  // [n,n-1,...,2,1]
 | 
						|
    }
 | 
						|
    count = bubbleSort(&nums);
 | 
						|
    std.debug.print("平方阶(冒泡排序)的操作数量 = {}\n", .{count});
 | 
						|
 | 
						|
    count = exponential(n);
 | 
						|
    std.debug.print("指数阶(循环实现)的操作数量 = {}\n", .{count});
 | 
						|
    count = expRecur(n);
 | 
						|
    std.debug.print("指数阶(递归实现)的操作数量 = {}\n", .{count});
 | 
						|
 | 
						|
    count = logarithmic(n);
 | 
						|
    std.debug.print("对数阶(循环实现)的操作数量 = {}\n", .{count});
 | 
						|
    count = logRecur(n);
 | 
						|
    std.debug.print("对数阶(递归实现)的操作数量 = {}\n", .{count});
 | 
						|
 | 
						|
    count = linearLogRecur(n);
 | 
						|
    std.debug.print("线性对数阶(递归实现)的操作数量 = {}\n", .{count});
 | 
						|
 | 
						|
    count = factorialRecur(n);
 | 
						|
    std.debug.print("阶乘阶(递归实现)的操作数量 = {}\n", .{count});
 | 
						|
 | 
						|
    _ = try std.io.getStdIn().reader().readByte();
 | 
						|
}
 | 
						|
 |