hello-algo/codes/c/chapter_computational_complexity/iteration.c
MwumLi 01732cc14f
feat: add c codes for iteration (#735)
Co-authored-by: Yudong Jin <krahets@163.com>
2023-09-13 03:13:55 +08:00

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/**
* File: iteration.c
* Created Time: 2023-09-09
* Author: Gonglja (glj0@outlook.com), MwumLi (mwumli@hotmail.com)
*/
#include "../utils/common.h"
/* for 循环 */
int forLoop(int n) {
int res = 0;
// 循环求和 1, 2, ..., n-1, n
for (int i = 1; i <= n; i++) {
res += i;
}
return res;
}
/* while 循环 */
int whileLoop(int n) {
int res = 0;
int i = 1; // 初始化条件变量
// 循环求和 1, 2, ..., n-1, n
while (i <= n) {
res += i;
i++; // 更新条件变量
}
return res;
}
/* while 循环(两次更新) */
int whileLoopII(int n) {
int res = 0;
int i = 1; // 初始化条件变量
// 循环求和 1, 4, ...
while (i <= n) {
res += i;
// 更新条件变量
i++;
i *= 2;
}
return res;
}
/* 双层 for 循环 */
char *nestedForLoop(int n) {
// n * n 为对应点数量,"(i, j), " 对应字符串长最大为 6+10*2加上最后一个空字符 \0 的额外空间
int size = n * n * 26 + 1;
char *res = malloc(size * sizeof(char));
// 循环 i = 1, 2, ..., n-1, n
for (int i = 1; i <= n; i++) {
// 循环 j = 1, 2, ..., n-1, n
for (int j = 1; j <= n; j++) {
char tmp[26];
snprintf(tmp, sizeof(tmp), "(%d, %d), ", i, j);
strncat(res, tmp, size - strlen(res) - 1);
}
}
return res;
}
/* Driver Code */
int main() {
int n = 5;
int res;
res = forLoop(n);
printf("\nfor 循环的求和结果 res = %d\n", res);
res = whileLoop(n);
printf("\nwhile 循环的求和结果 res = %d\n", res);
res = whileLoopII(n);
printf("\nwhile 循环(两次更新)求和结果 res = %d\n", res);
char *resStr = nestedForLoop(n);
printf("\n双层 for 循环的遍历结果 %s\r\n", resStr);
free(resStr);
return 0;
}