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180 lines
4.3 KiB
C
180 lines
4.3 KiB
C
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/**
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* File: time_complexity.c
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* Created Time: 2023-01-03
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* Author: codingonion (coderonion@gmail.com)
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*/
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#include "../utils/common.h"
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/* 常數階 */
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int constant(int n) {
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int count = 0;
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int size = 100000;
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int i = 0;
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for (int i = 0; i < size; i++) {
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count++;
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}
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return count;
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}
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/* 線性階 */
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int linear(int n) {
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int count = 0;
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for (int i = 0; i < n; i++) {
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count++;
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}
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return count;
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}
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/* 線性階(走訪陣列) */
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int arrayTraversal(int *nums, int n) {
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int count = 0;
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// 迴圈次數與陣列長度成正比
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for (int i = 0; i < n; i++) {
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count++;
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}
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return count;
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}
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/* 平方階 */
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int quadratic(int n) {
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int count = 0;
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// 迴圈次數與資料大小 n 成平方關係
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) {
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count++;
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}
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}
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return count;
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}
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/* 平方階(泡沫排序) */
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int bubbleSort(int *nums, int n) {
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int count = 0; // 計數器
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// 外迴圈:未排序區間為 [0, i]
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for (int i = n - 1; i > 0; i--) {
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// 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
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for (int j = 0; j < i; j++) {
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if (nums[j] > nums[j + 1]) {
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// 交換 nums[j] 與 nums[j + 1]
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int 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; // 元素交換包含 3 個單元操作
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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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/* 指數階(迴圈實現) */
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int exponential(int n) {
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int count = 0;
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int bas = 1;
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// 細胞每輪一分為二,形成數列 1, 2, 4, 8, ..., 2^(n-1)
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < bas; j++) {
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count++;
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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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/* 指數階(遞迴實現) */
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int expRecur(int n) {
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if (n == 1)
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return 1;
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return expRecur(n - 1) + expRecur(n - 1) + 1;
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}
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/* 對數階(迴圈實現) */
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int logarithmic(int n) {
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int count = 0;
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while (n > 1) {
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n = n / 2;
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count++;
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}
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return count;
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}
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/* 對數階(遞迴實現) */
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int logRecur(int n) {
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if (n <= 1)
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return 0;
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return logRecur(n / 2) + 1;
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}
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/* 線性對數階 */
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int linearLogRecur(int n) {
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if (n <= 1)
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return 1;
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int count = linearLogRecur(n / 2) + linearLogRecur(n / 2);
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for (int i = 0; i < n; i++) {
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count++;
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}
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return count;
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}
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/* 階乘階(遞迴實現) */
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int factorialRecur(int n) {
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if (n == 0)
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return 1;
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int count = 0;
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for (int i = 0; i < n; i++) {
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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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int main(int argc, char *argv[]) {
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// 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
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int n = 8;
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printf("輸入資料大小 n = %d\n", n);
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int count = constant(n);
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printf("常數階的操作數量 = %d\n", count);
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count = linear(n);
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printf("線性階的操作數量 = %d\n", count);
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// 分配堆積區記憶體(建立一維可變長陣列:陣列中元素數量為 n ,元素型別為 int )
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int *nums = (int *)malloc(n * sizeof(int));
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count = arrayTraversal(nums, n);
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printf("線性階(走訪陣列)的操作數量 = %d\n", count);
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count = quadratic(n);
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printf("平方階的操作數量 = %d\n", count);
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for (int i = 0; i < n; i++) {
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nums[i] = n - i; // [n,n-1,...,2,1]
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}
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count = bubbleSort(nums, n);
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printf("平方階(泡沫排序)的操作數量 = %d\n", count);
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count = exponential(n);
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printf("指數階(迴圈實現)的操作數量 = %d\n", count);
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count = expRecur(n);
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printf("指數階(遞迴實現)的操作數量 = %d\n", count);
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count = logarithmic(n);
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printf("對數階(迴圈實現)的操作數量 = %d\n", count);
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count = logRecur(n);
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printf("對數階(遞迴實現)的操作數量 = %d\n", count);
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count = linearLogRecur(n);
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printf("線性對數階(遞迴實現)的操作數量 = %d\n", count);
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count = factorialRecur(n);
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printf("階乘階(遞迴實現)的操作數量 = %d\n", count);
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// 釋放堆積區記憶體
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if (nums != NULL) {
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free(nums);
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nums = NULL;
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}
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getchar();
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return 0;
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}
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