1. Add build script for Java.

2. Add height limitation for code blocks in extra.css.
3. Fix "节点" to "结点".
This commit is contained in:
krahets
2023-02-07 04:43:52 +08:00
parent b14568151c
commit ecbf2d1560
54 changed files with 457 additions and 1633 deletions

View File

@ -795,14 +795,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 常数阶 */
int constant(int n) {
int count = 0;
int size = 100000;
for (int i = 0; i < size; i++)
count++;
return count;
}
[class]{time_complexity}-[func]{constant}
```
=== "C++"
@ -928,13 +921,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 线性阶 */
int linear(int n) {
int count = 0;
for (int i = 0; i < n; i++)
count++;
return count;
}
[class]{time_complexity}-[func]{linear}
```
=== "C++"
@ -1052,15 +1039,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 线性阶(遍历数组) */
int arrayTraversal(int[] nums) {
int count = 0;
// 循环次数与数组长度成正比
for (int num : nums) {
count++;
}
return count;
}
[class]{time_complexity}-[func]{arrayTraversal}
```
=== "C++"
@ -1190,17 +1169,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 平方阶 */
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++) {
count++;
}
}
return count;
}
[class]{time_complexity}-[func]{quadratic}
```
=== "C++"
@ -1355,24 +1324,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 平方阶(冒泡排序) */
int bubbleSort(int[] nums) {
int count = 0; // 计数器
// 外循环:待排序元素数量为 n-1, n-2, ..., 1
for (int i = nums.length - 1; i > 0; i--) {
// 内循环:冒泡操作
for (int j = 0; j < i; j++) {
if (nums[j] > nums[j + 1]) {
// 交换 nums[j] 与 nums[j + 1]
int tmp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = tmp;
count += 3; // 元素交换包含 3 个单元操作
}
}
}
return count;
}
[class]{time_complexity}-[func]{bubbleSort}
```
=== "C++"
@ -1585,19 +1537,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 指数阶(循环实现) */
int exponential(int n) {
int count = 0, base = 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 < base; j++) {
count++;
}
base *= 2;
}
// count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
return count;
}
[class]{time_complexity}-[func]{exponential}
```
=== "C++"
@ -1770,11 +1710,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 指数阶(递归实现) */
int expRecur(int n) {
if (n == 1) return 1;
return expRecur(n - 1) + expRecur(n - 1) + 1;
}
[class]{time_complexity}-[func]{expRecur}
```
=== "C++"
@ -1880,15 +1816,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 对数阶(循环实现) */
int logarithmic(float n) {
int count = 0;
while (n > 1) {
n = n / 2;
count++;
}
return count;
}
[class]{time_complexity}-[func]{logarithmic}
```
=== "C++"
@ -2024,11 +1952,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 对数阶(递归实现) */
int logRecur(float n) {
if (n <= 1) return 0;
return logRecur(n / 2) + 1;
}
[class]{time_complexity}-[func]{logRecur}
```
=== "C++"
@ -2132,16 +2056,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 线性对数阶 */
int linearLogRecur(float n) {
if (n <= 1) return 1;
int count = linearLogRecur(n / 2) +
linearLogRecur(n / 2);
for (int i = 0; i < n; i++) {
count++;
}
return count;
}
[class]{time_complexity}-[func]{linearLogRecur}
```
=== "C++"
@ -2292,16 +2207,7 @@ $$
=== "Java"
```java title="time_complexity.java"
/* 阶乘阶(递归实现) */
int factorialRecur(int n) {
if (n == 0) return 1;
int count = 0;
// 从 1 个分裂出 n 个
for (int i = 0; i < n; i++) {
count += factorialRecur(n - 1);
}
return count;
}
[class]{time_complexity}-[func]{factorialRecur}
```
=== "C++"
@ -2452,35 +2358,9 @@ $$
=== "Java"
```java title="worst_best_time_complexity.java"
public class worst_best_time_complexity {
/* 生成一个数组,元素为 { 1, 2, ..., n },顺序被打乱 */
int[] randomNumbers(int n) {
Integer[] nums = new Integer[n];
// 生成数组 nums = { 1, 2, 3, ..., n }
for (int i = 0; i < n; i++) {
nums[i] = i + 1;
}
// 随机打乱数组元素
Collections.shuffle(Arrays.asList(nums));
// Integer[] -> int[]
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nums[i];
}
return res;
}
[class]{worst_best_time_complexity}-[func]{randomNumbers}
/* 查找数组 nums 中数字 1 所在索引 */
int findOne(int[] nums) {
for (int i = 0; i < nums.length; i++) {
// 当元素 1 在数组头部时,达到最佳时间复杂度 O(1)
// 当元素 1 在数组尾部时,达到最差时间复杂度 O(n)
if (nums[i] == 1)
return i;
}
return -1;
}
}
[class]{worst_best_time_complexity}-[func]{findOne}
```
=== "C++"