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(PR #204) update c code and doc for time_complexity

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sjinzh
2023-01-05 08:39:09 +08:00
parent 9e4a5fd6d8 fd3eaaf3fd
commit ea867eadac
23 changed files with 447 additions and 172 deletions
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33
docs/chapter_computational_complexity/space_time_tradeoff.md
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@ -149,6 +149,22 @@ comments: true
}
```
=== "Swift"
```swift title="leetcode_two_sum.swift"
func twoSumBruteForce(nums: [Int], target: Int) -> [Int] {
// 两层循环，时间复杂度 O(n^2)
for i in nums.indices.dropLast() {
for j in nums.indices.dropFirst(i + 1) {
if nums[i] + nums[j] == target {
return [i, j]
}
}
}
return [0]
}
```
### 方法二：辅助哈希表
时间复杂度 $O(N)$ ，空间复杂度 $O(N)$ ，属于「空间换时间」。
@ -294,3 +310,20 @@ comments: true
}
}
```
=== "Swift"
```swift title="leetcode_two_sum.swift"
func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
// 辅助哈希表，空间复杂度 O(n)
var dic: [Int: Int] = [:]
// 单层循环，时间复杂度 O(n)
for i in nums.indices {
if let j = dic[target - nums[i]] {
return [j, i]
}
dic[nums[i]] = i
}
return [0]
}
```

111
docs/chapter_computational_complexity/time_complexity.md
View File

@ -92,8 +92,7 @@ $$
```c title=""
// 在某运行平台下
void algorithm(int n)
{
void algorithm(int n) {
int a = 2; // 1 ns
a = a + 1; // 1 ns
a = a * 2; // 10 ns
@ -243,20 +242,17 @@ $$
```c title=""
// 算法 A 时间复杂度：常数阶
void algorithm_A(int n)
{
void algorithm_A(int n) {
printf("%d", 0);
}
// 算法 B 时间复杂度：线性阶
void algorithm_B(int n)
{
void algorithm_B(int n) {
for (int i = 0; i < n; i++) {
printf("%d", 0);
}
}
// 算法 C 时间复杂度：常数阶
void algorithm_C(int n)
{
void algorithm_C(int n) {
for (int i = 0; i < 1000000; i++) {
printf("%d", 0);
}
@ -401,8 +397,7 @@ $$
=== "C"
```c title=""
void algorithm(int n)
{
void algorithm(int n) {
int a = 1; // +1
a = a + 1; // +1
a = a * 2; // +1
@ -579,8 +574,7 @@ $$
=== "C"
```c title=""
void algorithm(int n)
{
void algorithm(int n) {
int a = 1; // +0（技巧 1）
a = a + n; // +0（技巧 1）
// +n（技巧 2）
@ -749,12 +743,11 @@ $$
```c title="time_complexity.c"
/* 常数阶 */
int constant(int n)
{
int constant(int n) {
int count = 0;
int size = 100000;
int i = 0;
for(int i = 0; i < size; i++){
for (int i = 0; i < size; i++) {
count ++;
}
return count;
@ -857,10 +850,9 @@ $$
```c title="time_complexity.c"
/* 线性阶 */
int linear(int n)
{
int linear(int n) {
int count = 0;
for(int i = 0; i < n; i++){
for (int i = 0; i < n; i++) {
count ++;
}
return count;
@ -969,11 +961,10 @@ $$
```c title="time_complexity.c"
/* 线性阶（遍历数组） */
int arrayTraversal(int *nums, int n)
{
int arrayTraversal(int *nums, int n) {
int count = 0;
// 循环次数与数组长度成正比
for(int i = 0; i < n; i++){
for (int i = 0; i < n; i++) {
count ++;
}
return count;
@ -1091,12 +1082,11 @@ $$
```c title="time_complexity.c"
/* 平方阶 */
int quadratic(int n)
{
int quadratic(int n) {
int count = 0;
// 循环次数与数组长度成平方关系
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
count ++;
}
}
@ -1253,14 +1243,12 @@ $$
```c title="time_complexity.c"
/* 平方阶（冒泡排序） */
int bubbleSort(int *nums, int n)
{
int bubbleSort(int *nums, int n) {
int count = 0; // 计数器
// 外循环：待排序元素数量为 n-1, n-2, ..., 1
for(int i = n - 1; i > 0; i--){
for (int i = n - 1; i > 0; i--) {
// 内循环：冒泡操作
for (int j = 0; j < i; j++)
{
for (int j = 0; j < i; j++) {
// 交换 nums[j] 与 nums[j + 1]
int tmp = nums[j];
nums[j] = nums[j + 1];
@ -1417,15 +1405,12 @@ $$
```c title="time_complexity.c"
/* 指数阶（循环实现） */
int exponential(int n)
{
int exponential(int n) {
int count = 0;
int bas = 1;
// cell 每轮一分为二，形成数列 1, 2, 4, 8, ..., 2^(n-1)
for (int i = 0; i < n; i++)
{
for (int j = 0; j < bas; j++)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < bas; j++) {
count++;
}
bas *= 2;
@ -1538,8 +1523,7 @@ $$
```c title="time_complexity.c"
/* 指数阶（递归实现） */
int expRecur(int n)
{
int expRecur(int n) {
if (n == 1) return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
@ -1646,11 +1630,9 @@ $$
```c title="time_complexity.c"
/* 对数阶（循环实现） */
int logarithmic(float n)
{
int logarithmic(float n) {
int count = 0;
while (n > 1)
{
while (n > 1) {
n = n / 2;
count++;
}
@ -1752,8 +1734,7 @@ $$
```c title="time_complexity.c"
/* 对数阶（递归实现） */
int logRecur(float n)
{
int logRecur(float n) {
if (n <= 1) return 0;
return logRecur(n / 2) + 1;
}
@ -1864,13 +1845,11 @@ $$
```c title="time_complexity.c"
/* 线性对数阶 */
int linearLogRecur(float n)
{
int linearLogRecur(float n) {
if (n <= 1) return 1;
int count = linearLogRecur(n / 2) +
linearLogRecur(n / 2);
for (int i = 0; i < n; i++)
{
for (int i = 0; i < n; i++) {
count ++;
}
return count;
@ -2000,12 +1979,10 @@ $$
```c title="time_complexity.c"
/* 阶乘阶（递归实现） */
int factorialRecur(int n)
{
int factorialRecur(int n) {
if (n == 0) return 1;
int count = 0;
for (int i = 0; i < n; i++)
{
for (int i = 0; i < n; i++) {
count += factorialRecur(n - 1);
}
return count;
@ -2226,18 +2203,15 @@ $$
```c title="worst_best_time_complexity.c"
/* 生成一个数组，元素为 { 1, 2, ..., n }，顺序被打乱 */
int *randomNumbers(int n)
{
int *randomNumbers(int n) {
// 分配堆区内存（创建一维可变长数组：数组中元素数量为n，元素类型为int）
int *nums = (int *)malloc(n * sizeof(int));
// 生成数组 nums = { 1, 2, 3, ..., n }
for (int i = 0; i < n; i++)
{
for (int i = 0; i < n; i++) {
nums[i] = i + 1;
}
// 随机打乱数组元素
for (int i = n - 1; i > 0; i--)
{
for (int i = n - 1; i > 0; i--) {
int j = rand() % (i + 1);
int temp = nums[i];
nums[i] = nums[j];
@ -2247,33 +2221,26 @@ $$
}
/* 查找数组 nums 中数字 1 所在索引 */
int findOne(int *nums, int n)
{
for (int i = 0; i < n; i++)
{
int findOne(int *nums, int n) {
for (int i = 0; i < n; i++) {
if (nums[i] == 1) return i;
}
return -1;
}
/* Driver Code */
int main(int argc, char *argv[])
{
int main(int argc, char *argv[]) {
// 初始化随机数种子
srand((unsigned int)time(NULL));
for (int i = 0; i < 10; i++)
{
for (int i = 0; i < 10; i++) {
int n = 100;
int *nums = randomNumbers(n);
int index = findOne(nums, n);
printf("\n数组 [ 1, 2, ..., n ] 被打乱后 = ");
for (int j = 0; j < n; j++)
{
printf("%d%s", nums[j], (j == n-1)? "" : "," );
}
printf("\n数字 1 的索引为 %d\n", index);
printArray(nums, n);
printf("数字 1 的索引为 %d\n", index);
// 释放堆区内存
if(nums != NULL){
if (nums != NULL) {
free(nums);
nums = NULL;
}