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3ea91bda99
* Use int instead of float for the example code of log time complexity * Bug fixes * Bug fixes
179 lines
4.6 KiB
Zig
179 lines
4.6 KiB
Zig
// File: time_complexity.zig
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// Created Time: 2022-12-28
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// Author: codingonion (coderonion@gmail.com)
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const std = @import("std");
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// 常数阶
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fn constant(n: i32) i32 {
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_ = n;
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var count: i32 = 0;
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const size: i32 = 100_000;
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var i: i32 = 0;
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while(i<size) : (i += 1) {
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count += 1;
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}
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return count;
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}
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// 线性阶
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fn linear(n: i32) i32 {
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var count: i32 = 0;
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var i: i32 = 0;
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while (i < n) : (i += 1) {
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count += 1;
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}
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return count;
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}
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// 线性阶(遍历数组)
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fn arrayTraversal(nums: []i32) i32 {
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var count: i32 = 0;
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// 循环次数与数组长度成正比
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for (nums) |_| {
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count += 1;
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}
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return count;
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}
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// 平方阶
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fn quadratic(n: i32) i32 {
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var count: i32 = 0;
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var i: i32 = 0;
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// 循环次数与数据大小 n 成平方关系
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while (i < n) : (i += 1) {
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var j: i32 = 0;
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while (j < n) : (j += 1) {
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count += 1;
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}
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}
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return count;
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}
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// 平方阶(冒泡排序)
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fn bubbleSort(nums: []i32) i32 {
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var count: i32 = 0; // 计数器
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// 外循环:未排序区间为 [0, i]
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var i: i32 = @as(i32, @intCast(nums.len)) - 1;
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while (i > 0) : (i -= 1) {
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var j: usize = 0;
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// 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
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while (j < i) : (j += 1) {
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if (nums[j] > nums[j + 1]) {
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// 交换 nums[j] 与 nums[j + 1]
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var tmp = nums[j];
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nums[j] = nums[j + 1];
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nums[j + 1] = tmp;
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count += 3; // 元素交换包含 3 个单元操作
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}
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}
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}
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return count;
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}
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// 指数阶(循环实现)
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fn exponential(n: i32) i32 {
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var count: i32 = 0;
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var bas: i32 = 1;
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var i: i32 = 0;
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// 细胞每轮一分为二,形成数列 1, 2, 4, 8, ..., 2^(n-1)
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while (i < n) : (i += 1) {
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var j: i32 = 0;
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while (j < bas) : (j += 1) {
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count += 1;
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}
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bas *= 2;
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}
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// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count;
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}
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// 指数阶(递归实现)
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fn expRecur(n: i32) i32 {
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if (n == 1) return 1;
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return expRecur(n - 1) + expRecur(n - 1) + 1;
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}
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// 对数阶(循环实现)
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fn logarithmic(n: i32) i32 {
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var count: i32 = 0;
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var n_var = n;
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while (n_var > 1)
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{
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n_var = n_var / 2;
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count +=1;
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}
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return count;
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}
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// 对数阶(递归实现)
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fn logRecur(n: i32) i32 {
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if (n <= 1) return 0;
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return logRecur(n / 2) + 1;
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}
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// 线性对数阶
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fn linearLogRecur(n: i32) i32 {
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if (n <= 1) return 1;
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var count: i32 = linearLogRecur(n / 2) + linearLogRecur(n / 2);
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var i: i32 = 0;
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while (i < n) : (i += 1) {
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count += 1;
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}
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return count;
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}
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// 阶乘阶(递归实现)
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fn factorialRecur(n: i32) i32 {
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if (n == 0) return 1;
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var count: i32 = 0;
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var i: i32 = 0;
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// 从 1 个分裂出 n 个
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while (i < n) : (i += 1) {
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count += factorialRecur(n - 1);
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}
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return count;
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}
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// Driver Code
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pub fn main() !void {
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// 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
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const n: i32 = 8;
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std.debug.print("输入数据大小 n = {}\n", .{n});
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var count = constant(n);
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std.debug.print("常数阶的操作数量 = {}\n", .{count});
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count = linear(n);
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std.debug.print("线性阶的操作数量 = {}\n", .{count});
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var nums = [_]i32{0}**n;
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count = arrayTraversal(&nums);
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std.debug.print("线性阶(遍历数组)的操作数量 = {}\n", .{count});
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count = quadratic(n);
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std.debug.print("平方阶的操作数量 = {}\n", .{count});
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for (&nums, 0..) |*num, i| {
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num.* = n - @as(i32, @intCast(i)); // [n,n-1,...,2,1]
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}
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count = bubbleSort(&nums);
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std.debug.print("平方阶(冒泡排序)的操作数量 = {}\n", .{count});
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count = exponential(n);
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std.debug.print("指数阶(循环实现)的操作数量 = {}\n", .{count});
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count = expRecur(n);
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std.debug.print("指数阶(递归实现)的操作数量 = {}\n", .{count});
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count = logarithmic(n);
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std.debug.print("对数阶(循环实现)的操作数量 = {}\n", .{count});
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count = logRecur(n);
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std.debug.print("对数阶(递归实现)的操作数量 = {}\n", .{count});
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count = linearLogRecur(n);
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std.debug.print("线性对数阶(递归实现)的操作数量 = {}\n", .{count});
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count = factorialRecur(n);
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std.debug.print("阶乘阶(递归实现)的操作数量 = {}\n", .{count});
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_ = try std.io.getStdIn().reader().readByte();
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}
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