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feat: add solutions to lc problem: No.2876 #1742

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Oct 3, 2023
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feat: add solutions to lc problem: No.2876 #1742

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152 changes: 149 additions & 3 deletions solution/2800-2899/2876.Count Visited Nodes in a Directed Graph/README.md
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -55,20 +55,81 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：基环树 + 遍历搜索**

我们可以用一个数组 $ans$ 记录每个节点的答案，用一个数组 $vis$ 记录每个节点的访问次序。

遍历每个节点 $i$，如果当前节点 $i$ 未被访问，我们就从节点 $i$ 开始遍历，此时有两种情况：

1. 如果遍历过程中，遇到了当前节点出发时走过的节点，那么此次遍历，一定是先走到了环内，然后沿着环走了一圈。对于环外的节点，其答案就是环的长度加上节点到环的距离；对于环内的节点，其答案就是环的长度。
1. 如果遍历过程中，遇到了此前节点出发时走过的节点，那么对于每个走过的节点，其答案就是当前节点到此节点的距离，加上此节点的答案。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $edges$ 的长度。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def countVisitedNodes(self, edges: List[int]) -> List[int]:
n = len(edges)
ans = [0] * n
vis = [0] * n
for i in range(n):
if not ans[i]:
cnt, j = 0, i
while not vis[j]:
cnt += 1
vis[j] = cnt
j = edges[j]
cycle, total = 0, cnt + ans[j]
if not ans[j]:
cycle = cnt - vis[j] + 1
total = cnt
j = i
while not ans[j]:
ans[j] = max(total, cycle)
total -= 1
j = edges[j]
return ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public int[] countVisitedNodes(List<Integer> edges) {
int n = edges.size();
int[] ans = new int[n];
int[] vis = new int[n];
for (int i = 0; i < n; ++i) {
if (ans[i] == 0) {
int cnt = 0, j = i;
while (vis[j] == 0) {
vis[j] = ++cnt;
j = edges.get(j);
}
int cycle = 0, total = cnt + ans[j];
if (ans[j] == 0) {
cycle = cnt - vis[j] + 1;
}
j = i;
while (ans[j] == 0) {
ans[j] = Math.max(total--, cycle);
j = edges.get(j);
}
}
}
return ans;
}
}
```

```java
class Solution {
void dfs(int curr, List<Integer> edges, int[] ans) {
Expand Down Expand Up @@ -106,19 +167,104 @@ class Solution {
return ans;
}
}

```

### **C++**

```cpp

class Solution {
public:
vector<int> countVisitedNodes(vector<int>& edges) {
int n = edges.size();
vector<int> ans(n), vis(n);
for (int i = 0; i < n; ++i) {
if (!ans[i]) {
int cnt = 0, j = i;
while (vis[j] == 0) {
vis[j] = ++cnt;
j = edges[j];
}
int cycle = 0, total = cnt + ans[j];
if (ans[j] == 0) {
cycle = cnt - vis[j] + 1;
}
j = i;
while (ans[j] == 0) {
ans[j] = max(total--, cycle);
j = edges[j];
}
}
}
return ans;
}
};
```

### **Go**

```go
func countVisitedNodes(edges []int) []int {
n := len(edges)
ans := make([]int, n)
vis := make([]int, n)
for i := range ans {
if ans[i] == 0 {
cnt, j := 0, i
for vis[j] == 0 {
cnt++
vis[j] = cnt
j = edges[j]
}
cycle, total := 0, cnt+ans[j]
if ans[j] == 0 {
cycle = cnt - vis[j] + 1
}
j = i
for ans[j] == 0 {
ans[j] = max(total, cycle)
total--
j = edges[j]
}
}
}
return ans
}

func max(a, b int) int {
if a > b {
return a
}
return b
}
```

### **TypeScript**

```ts
function countVisitedNodes(edges: number[]): number[] {
const n = edges.length;
const ans: number[] = Array(n).fill(0);
const vis: number[] = Array(n).fill(0);
for (let i = 0; i < n; ++i) {
if (ans[i] === 0) {
let [cnt, j] = [0, i];
while (vis[j] === 0) {
vis[j] = ++cnt;
j = edges[j];
}
let [cycle, total] = [0, cnt + ans[j]];
if (ans[j] === 0) {
cycle = cnt - vis[j] + 1;
}
j = i;
while (ans[j] === 0) {
ans[j] = Math.max(total--, cycle);
j = edges[j];
}
}
}
return ans;
}
```

### **...**
Expand Down
152 changes: 149 additions & 3 deletions solution/2800-2899/2876.Count Visited Nodes in a Directed Graph/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -49,16 +49,77 @@

## Solutions

**Solution 1: Basic Tree + Traversal**

We can use an array $ans$ to record the answer for each node, and an array $vis$ to record the visit order for each node.

For each node $i$, if it has not been visited yet, we start traversing from node $i$. There are two cases:

- If we encounter a node that has been visited before during the traversal, then we must have first entered the cycle and then walked around the cycle. For nodes outside the cycle, their answer is the length of the cycle plus the distance from the node to the cycle; for nodes inside the cycle, their answer is the length of the cycle.
- If we encounter a node that has been visited before during the traversal, then for each visited node, its answer is the distance from the current node to this node plus the answer of this node.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array edges.

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def countVisitedNodes(self, edges: List[int]) -> List[int]:
n = len(edges)
ans = [0] * n
vis = [0] * n
for i in range(n):
if not ans[i]:
cnt, j = 0, i
while not vis[j]:
cnt += 1
vis[j] = cnt
j = edges[j]
cycle, total = 0, cnt + ans[j]
if not ans[j]:
cycle = cnt - vis[j] + 1
total = cnt
j = i
while not ans[j]:
ans[j] = max(total, cycle)
total -= 1
j = edges[j]
return ans
```

### **Java**

```java
class Solution {
public int[] countVisitedNodes(List<Integer> edges) {
int n = edges.size();
int[] ans = new int[n];
int[] vis = new int[n];
for (int i = 0; i < n; ++i) {
if (ans[i] == 0) {
int cnt = 0, j = i;
while (vis[j] == 0) {
vis[j] = ++cnt;
j = edges.get(j);
}
int cycle = 0, total = cnt + ans[j];
if (ans[j] == 0) {
cycle = cnt - vis[j] + 1;
}
j = i;
while (ans[j] == 0) {
ans[j] = Math.max(total--, cycle);
j = edges.get(j);
}
}
}
return ans;
}
}
```

```java
class Solution {
void dfs(int curr, List<Integer> edges, int[] ans) {
Expand Down Expand Up @@ -96,19 +157,104 @@ class Solution {
return ans;
}
}

```

### **C++**

```cpp

class Solution {
public:
vector<int> countVisitedNodes(vector<int>& edges) {
int n = edges.size();
vector<int> ans(n), vis(n);
for (int i = 0; i < n; ++i) {
if (!ans[i]) {
int cnt = 0, j = i;
while (vis[j] == 0) {
vis[j] = ++cnt;
j = edges[j];
}
int cycle = 0, total = cnt + ans[j];
if (ans[j] == 0) {
cycle = cnt - vis[j] + 1;
}
j = i;
while (ans[j] == 0) {
ans[j] = max(total--, cycle);
j = edges[j];
}
}
}
return ans;
}
};
```

### **Go**

```go
func countVisitedNodes(edges []int) []int {
n := len(edges)
ans := make([]int, n)
vis := make([]int, n)
for i := range ans {
if ans[i] == 0 {
cnt, j := 0, i
for vis[j] == 0 {
cnt++
vis[j] = cnt
j = edges[j]
}
cycle, total := 0, cnt+ans[j]
if ans[j] == 0 {
cycle = cnt - vis[j] + 1
}
j = i
for ans[j] == 0 {
ans[j] = max(total, cycle)
total--
j = edges[j]
}
}
}
return ans
}

func max(a, b int) int {
if a > b {
return a
}
return b
}
```

### **TypeScript**

```ts
function countVisitedNodes(edges: number[]): number[] {
const n = edges.length;
const ans: number[] = Array(n).fill(0);
const vis: number[] = Array(n).fill(0);
for (let i = 0; i < n; ++i) {
if (ans[i] === 0) {
let [cnt, j] = [0, i];
while (vis[j] === 0) {
vis[j] = ++cnt;
j = edges[j];
}
let [cycle, total] = [0, cnt + ans[j]];
if (ans[j] === 0) {
cycle = cnt - vis[j] + 1;
}
j = i;
while (ans[j] === 0) {
ans[j] = Math.max(total--, cycle);
j = edges[j];
}
}
}
return ans;
}
```

### **...**
Expand Down
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