mirror of
https://github.com/krahets/hello-algo.git
synced 2026-07-12 15:36:05 +00:00
build
This commit is contained in:
@@ -978,7 +978,8 @@ $$
|
||||
/* 线性阶(递归实现) */
|
||||
void linearRecur(int n) {
|
||||
System.out.println("递归 n = " + n);
|
||||
if (n == 1) return;
|
||||
if (n == 1)
|
||||
return;
|
||||
linearRecur(n - 1);
|
||||
}
|
||||
```
|
||||
@@ -989,7 +990,8 @@ $$
|
||||
/* 线性阶(递归实现) */
|
||||
void linearRecur(int n) {
|
||||
cout << "递归 n = " << n << endl;
|
||||
if (n == 1) return;
|
||||
if (n == 1)
|
||||
return;
|
||||
linearRecur(n - 1);
|
||||
}
|
||||
```
|
||||
@@ -1254,7 +1256,8 @@ $$
|
||||
```java title="space_complexity.java"
|
||||
/* 平方阶(递归实现) */
|
||||
int quadraticRecur(int n) {
|
||||
if (n <= 0) return 0;
|
||||
if (n <= 0)
|
||||
return 0;
|
||||
// 数组 nums 长度为 n, n-1, ..., 2, 1
|
||||
int[] nums = new int[n];
|
||||
System.out.println("递归 n = " + n + " 中的 nums 长度 = " + nums.length);
|
||||
@@ -1267,7 +1270,8 @@ $$
|
||||
```cpp title="space_complexity.cpp"
|
||||
/* 平方阶(递归实现) */
|
||||
int quadraticRecur(int n) {
|
||||
if (n <= 0) return 0;
|
||||
if (n <= 0)
|
||||
return 0;
|
||||
vector<int> nums(n);
|
||||
cout << "递归 n = " << n << " 中的 nums 长度 = " << nums.size() << endl;
|
||||
return quadraticRecur(n - 1);
|
||||
@@ -1384,7 +1388,8 @@ $$
|
||||
```java title="space_complexity.java"
|
||||
/* 指数阶(建立满二叉树) */
|
||||
TreeNode buildTree(int n) {
|
||||
if (n == 0) return null;
|
||||
if (n == 0)
|
||||
return null;
|
||||
TreeNode root = new TreeNode(0);
|
||||
root.left = buildTree(n - 1);
|
||||
root.right = buildTree(n - 1);
|
||||
@@ -1396,9 +1401,10 @@ $$
|
||||
|
||||
```cpp title="space_complexity.cpp"
|
||||
/* 指数阶(建立满二叉树) */
|
||||
TreeNode* buildTree(int n) {
|
||||
if (n == 0) return nullptr;
|
||||
TreeNode* root = new TreeNode(0);
|
||||
TreeNode *buildTree(int n) {
|
||||
if (n == 0)
|
||||
return nullptr;
|
||||
TreeNode *root = new TreeNode(0);
|
||||
root->left = buildTree(n - 1);
|
||||
root->right = buildTree(n - 1);
|
||||
return root;
|
||||
|
||||
Reference in New Issue
Block a user