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feat: add solutions to lc problem: No.0277
No.0277.Find the Celebrity
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‎solution/0200-0299/0277.Find the Celebrity/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：O(n) 遍历**
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经过验证，若暴力遍历，调用 $O(n^2)$ 次 $knows$ 方法，会报 TLE 错误。因此，我们需要寻找更优的解法。
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要找出 $n$ 个人中的名人，题目给我们的关键信息是：1. 名人不认识其他所有人；2. 其他所有人都认识名人。
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那么，我们初始时假定名人 $ans=0$。然后在 $[1,n)$ 范围内遍历 $i$，若 $ans$ 认识 $i$，说明 $ans$ 不是我们要找的名人，此时我们可以直接将 $ans$ 更新为 $i$。
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为什么呢？我们来举个实际的例子。
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```bash
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ans = 0
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for i in [1,n) {
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if (ans knows i) {
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ans = i
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}
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}
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ans = 0
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ans not knows 1
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ans not knows 2
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ans knows 3
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ans = 3
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ans not knows 4
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ans not knows 5
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ans not knows 6
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ans = 6
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```
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$ans$ 认识 $3$，说明 $ans$ 不是名人（$0$ 不是名人），那么名人会是 $1$ 或者 $2$ 吗？不会！因为若 $1$ 或者 $2$ 是名人，那么 $0$ 应该认识 $1$ 或者 $2$ 才对，与前面的例子冲突。因此，我们可以直接将 $ans$ 更新为 $i$。
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我们找出 $ans$ 之后，接下来再遍历一遍，判断 $ans$ 是否满足名人的条件。若不满足，返回 $-1$。
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否则遍历结束，返回 $ans$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# The knows API is already defined for you.
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# return a bool, whether a knows b
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# def knows(a: int, b: int) -> bool:
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class Solution:
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def findCelebrity(self, n: int) -> int:
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ans = 0
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for i in range(1, n):
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if knows(ans, i):
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ans = i
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for i in range(n):
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if ans != i:
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if knows(ans, i) or not knows(i, ans):
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return -1
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/* The knows API is defined in the parent class Relation.
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boolean knows(int a, int b); */
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public class Solution extends Relation {
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public int findCelebrity(int n) {
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int ans = 0;
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for (int i = 1; i < n; ++i) {
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if (knows(ans, i)) {
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ans = i;
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}
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}
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for (int i = 0; i < n; ++i) {
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if (ans != i) {
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if (knows(ans, i) || !knows(i, ans)) {
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return -1;
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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/* The knows API is defined for you.
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bool knows(int a, int b); */
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class Solution {
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public:
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int findCelebrity(int n) {
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int ans = 0;
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for (int i = 1; i < n; ++i) {
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if (knows(ans, i)) {
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ans = i;
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}
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}
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for (int i = 0; i < n; ++i) {
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if (ans != i) {
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if (knows(ans, i) || !knows(i, ans)) {
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return -1;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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/**
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* The knows API is already defined for you.
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* knows := func(a int, b int) bool
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*/
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func solution(knows func(a int, b int) bool) func(n int) int {
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return func(n int) int {
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ans := 0
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for i := 1; i < n; i++ {
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if knows(ans, i) {
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ans = i
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}
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}
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for i := 0; i < n; i++ {
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if ans != i {
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if knows(ans, i) || !knows(i, ans) {
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return -1
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}
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}
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}
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return ans
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}
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}
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```
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### **...**

‎solution/0200-0299/0277.Find the Celebrity/README_EN.md

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### **Python3**
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```python
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# The knows API is already defined for you.
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# return a bool, whether a knows b
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# def knows(a: int, b: int) -> bool:
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class Solution:
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def findCelebrity(self, n: int) -> int:
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ans = 0
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for i in range(1, n):
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if knows(ans, i):
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ans = i
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for i in range(n):
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if ans != i:
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if knows(ans, i) or not knows(i, ans):
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return -1
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return ans
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```
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### **Java**
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```java
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/* The knows API is defined in the parent class Relation.
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boolean knows(int a, int b); */
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public class Solution extends Relation {
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public int findCelebrity(int n) {
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int ans = 0;
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for (int i = 1; i < n; ++i) {
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if (knows(ans, i)) {
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ans = i;
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}
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}
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for (int i = 0; i < n; ++i) {
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if (ans != i) {
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if (knows(ans, i) || !knows(i, ans)) {
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return -1;
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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/* The knows API is defined for you.
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bool knows(int a, int b); */
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class Solution {
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public:
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int findCelebrity(int n) {
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int ans = 0;
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for (int i = 1; i < n; ++i) {
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if (knows(ans, i)) {
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ans = i;
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}
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}
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for (int i = 0; i < n; ++i) {
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if (ans != i) {
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if (knows(ans, i) || !knows(i, ans)) {
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return -1;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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/**
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* The knows API is already defined for you.
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* knows := func(a int, b int) bool
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*/
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func solution(knows func(a int, b int) bool) func(n int) int {
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return func(n int) int {
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ans := 0
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for i := 1; i < n; i++ {
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if knows(ans, i) {
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ans = i
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}
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}
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for i := 0; i < n; i++ {
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if ans != i {
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if knows(ans, i) || !knows(i, ans) {
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return -1
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}
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}
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}
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return ans
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}
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}
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```
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### **...**

‎solution/0200-0299/0277.Find the Celebrity/Solution.cpp

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/* The knows API is defined for you.
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bool knows(int a, int b); */
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class Solution {
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public:
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int findCelebrity(int n) {
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int ans = 0;
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for (int i = 1; i < n; ++i) {
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if (knows(ans, i)) {
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ans = i;
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}
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}
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for (int i = 0; i < n; ++i) {
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if (ans != i) {
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if (knows(ans, i) || !knows(i, ans)) {
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return -1;
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}
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}
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}
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return ans;
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}
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};

‎solution/0200-0299/0277.Find the Celebrity/Solution.go

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/**
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* The knows API is already defined for you.
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* knows := func(a int, b int) bool
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*/
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func solution(knows func(a int, b int) bool) func(n int) int {
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return func(n int) int {
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ans := 0
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for i := 1; i < n; i++ {
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if knows(ans, i) {
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ans = i
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}
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}
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for i := 0; i < n; i++ {
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if ans != i {
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if knows(ans, i) || !knows(i, ans) {
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return -1
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}
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}
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}
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return ans
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}
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}

‎solution/0200-0299/0277.Find the Celebrity/Solution.java

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/* The knows API is defined in the parent class Relation.
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boolean knows(int a, int b); */
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public class Solution extends Relation {
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public int findCelebrity(int n) {
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int ans = 0;
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for (int i = 1; i < n; ++i) {
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if (knows(ans, i)) {
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ans = i;
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}
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}
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for (int i = 0; i < n; ++i) {
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if (ans != i) {
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if (knows(ans, i) || !knows(i, ans)) {
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return -1;
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}
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}
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}
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return ans;
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}
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}

‎solution/0200-0299/0277.Find the Celebrity/Solution.py

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# The knows API is already defined for you.
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# return a bool, whether a knows b
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# def knows(a: int, b: int) -> bool:
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class Solution:
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def findCelebrity(self, n: int) -> int:
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ans = 0
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for i in range(1, n):
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if knows(ans, i):
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ans = i
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for i in range(n):
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if ans != i:
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if knows(ans, i) or not knows(i, ans):
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return -1
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return ans

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