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Commit ab7c381

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check in source code
1 parent ccbd9b6 commit ab7c381

9 files changed

+795
-0
lines changed
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using System;
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using System.Collections.Generic;
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using System.Linq;
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using System.Text;
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using System.Threading.Tasks;
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namespace Leetcode277_FindCelebrity
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{
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/* Leetcode 277: find celebrity
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*
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* Suppose you are at a party with n people (labeled from 0 to n - 1) and among them,
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* there may exist one celebrity. The definition of a celebrity is that all the other
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* n - 1 people know him/her but he/she does not know any of them.
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Now you want to find out who the celebrity is or verify that there is not one. The
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* only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?"
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* to get information of whether A knows B. You need to find out the celebrity (or
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* verify there is not one) by asking as few questions as possible (in the asymptotic
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* sense).
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You are given a helper function bool knows(a, b) which tells you whether A knows B.
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* Implement a function int findCelebrity(n), your function should minimize the number
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* of calls to knows.
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Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's
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* label if there is a celebrity in the party. If there is no celebrity, return -1.
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*
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* source code is based on
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* http://www.cnblogs.com/grandyang/p/5310649.html
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*/
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class Program
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{
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static void Main(string[] args)
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{
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var celebrity = FindCelebrity(100);
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}
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/// <summary>
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/// my job is to find O(N) solution, N is number of candidates
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/// </summary>
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/// <param name="n"></param>
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/// <returns></returns>
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public static int FindCelebrity(int n)
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{
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if (n <= 0)
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{
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return -1;
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}
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for (int i = 0, j = 0; i < n; ++i)
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{
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for (j = 0; j < n; ++j)
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{
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if (i != j && (knows(i, j) || !knows(j, i))) break;
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}
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if (j == n) return i;
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}
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return -1;
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}
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/// <summary>
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/// assume that there are more than 10 people, and make 10th people as a celebrity.
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/// </summary>
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/// <param name="a"></param>
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/// <param name="b"></param>
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/// <returns></returns>
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private static bool knows(int a, int b)
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{
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if (a == b)
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return true;
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if (a == 10)
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return false;
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return true; // false
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}
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}
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}
Original file line numberDiff line numberDiff line change
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using System;
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using System.Collections.Generic;
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using System.Linq;
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using System.Text;
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using System.Threading.Tasks;
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namespace Leetcode277_FindCelebrity
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{
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/* Leetcode 277: find celebrity
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*
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* Suppose you are at a party with n people (labeled from 0 to n - 1) and among them,
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* there may exist one celebrity. The definition of a celebrity is that all the other
13+
* n - 1 people know him/her but he/she does not know any of them.
14+
15+
Now you want to find out who the celebrity is or verify that there is not one. The
16+
* only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?"
17+
* to get information of whether A knows B. You need to find out the celebrity (or
18+
* verify there is not one) by asking as few questions as possible (in the asymptotic
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* sense).
20+
21+
You are given a helper function bool knows(a, b) which tells you whether A knows B.
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* Implement a function int findCelebrity(n), your function should minimize the number
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* of calls to knows.
24+
25+
Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's
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* label if there is a celebrity in the party. If there is no celebrity, return -1.
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*
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* source code is based on
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* http://www.cnblogs.com/grandyang/p/5310649.html
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*/
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class Program
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{
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static void Main(string[] args)
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{
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var celebrity = FindCelebrity(100);
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}
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/// <summary>
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/// my job is to find O(N) solution, N is number of candidates
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/// </summary>
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/// <param name="n"></param>
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/// <returns></returns>
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public static int FindCelebrity(int n)
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{
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if (n <= 0)
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{
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return -1;
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}
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// find the candidate first
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int candidate = 0;
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for (int i = 0; i < n; ++i)
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{
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if (candidate != i && knows(candidate, i))
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{
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candidate = i;
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}
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}
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// verify candidate is the celebrity
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for (int i = 0; i < n; ++i)
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{
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if (candidate != i && (knows(candidate, i) || !knows(i, candidate)))
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{
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return -1;
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}
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}
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return candidate;
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}
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/// <summary>
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/// assume that there are more than 10 people, and make 10th people as a celebrity.
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/// </summary>
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/// <param name="a"></param>
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/// <param name="b"></param>
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/// <returns></returns>
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private static bool knows(int a, int b)
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{
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if (a == b)
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return true;
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if (a == 10)
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return false;
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return true; // false
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}
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}
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}
Lines changed: 96 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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using System;
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using System.Collections.Generic;
3+
using System.Linq;
4+
using System.Text;
5+
using System.Threading.Tasks;
6+
7+
namespace Leetcode277_FindCelebrity
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{
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/* Leetcode 277: find celebrity
10+
*
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* Suppose you are at a party with n people (labeled from 0 to n - 1) and among them,
12+
* there may exist one celebrity. The definition of a celebrity is that all the other
13+
* n - 1 people know him/her but he/she does not know any of them.
14+
15+
Now you want to find out who the celebrity is or verify that there is not one. The
16+
* only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?"
17+
* to get information of whether A knows B. You need to find out the celebrity (or
18+
* verify there is not one) by asking as few questions as possible (in the asymptotic
19+
* sense).
20+
21+
You are given a helper function bool knows(a, b) which tells you whether A knows B.
22+
* Implement a function int findCelebrity(n), your function should minimize the number
23+
* of calls to knows.
24+
25+
Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's
26+
* label if there is a celebrity in the party. If there is no celebrity, return -1.
27+
*
28+
* source code is based on
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* http://www.cnblogs.com/grandyang/p/5310649.html
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*/
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class Program
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{
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static void Main(string[] args)
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{
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var celebrity = FindCelebrity(100);
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}
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/// <summary>
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/// my job is to find O(N) solution, N is number of candidates
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/// </summary>
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/// <param name="n"></param>
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/// <returns></returns>
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public static int FindCelebrity(int n)
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{
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if (n <= 0)
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return -1;
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var candidates = new bool[n];
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for (int i = 0; i < n; i++)
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candidates[i] = true;
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for (int i = 0; i < n; ++i)
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{
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for (int j = 0; j < n; ++j)
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{
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if (candidates[i] && i != j)
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{
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if (knows(i, j) || !knows(j, i)) // i knows j, or j does not know i
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{
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candidates[i] = false;
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break;
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}
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else
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{
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candidates[j] = false;
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}
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}
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}
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if (candidates[i])
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{
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return i;
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}
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}
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return -1;
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}
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/// <summary>
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/// assume that there are more than 10 people, and make 10th people as a celebrity.
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/// </summary>
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/// <param name="a"></param>
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/// <param name="b"></param>
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/// <returns></returns>
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private static bool knows(int a, int b)
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{
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if (a == b)
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return true;
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if (a == 10)
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return false;
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return true; // false
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}
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}
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}
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,80 @@
1+
using System;
2+
using System.Collections.Generic;
3+
using System.Linq;
4+
using System.Text;
5+
using System.Threading.Tasks;
6+
7+
namespace Leetcode277_FindCelebrity
8+
{
9+
/* Leetcode 277: find celebrity
10+
*
11+
* Suppose you are at a party with n people (labeled from 0 to n - 1) and among them,
12+
* there may exist one celebrity. The definition of a celebrity is that all the other
13+
* n - 1 people know him/her but he/she does not know any of them.
14+
15+
Now you want to find out who the celebrity is or verify that there is not one. The
16+
* only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?"
17+
* to get information of whether A knows B. You need to find out the celebrity (or
18+
* verify there is not one) by asking as few questions as possible (in the asymptotic
19+
* sense).
20+
21+
You are given a helper function bool knows(a, b) which tells you whether A knows B.
22+
* Implement a function int findCelebrity(n), your function should minimize the number
23+
* of calls to knows.
24+
25+
Note: There will be exactly one celebrity if he/she is in the party. Return the celebrity's
26+
* label if there is a celebrity in the party. If there is no celebrity, return -1.
27+
*
28+
* source code is based on
29+
* http://www.cnblogs.com/grandyang/p/5310649.html
30+
*/
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class Program
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{
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static void Main(string[] args)
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{
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var celebrity = FindCelebrity(100);
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}
37+
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/// <summary>
39+
/// my job is to find O(N) solution, N is number of candidates
40+
/// </summary>
41+
/// <param name="n"></param>
42+
/// <returns></returns>
43+
public static int FindCelebrity(int n)
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{
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if (n <= 0)
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{
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return -1;
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}
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for (int i = 0, j = 0; i < n; ++i)
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{
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for (j = 0; j < n; ++j)
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{
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if (i != j && (knows(i, j) || !knows(j, i))) break;
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}
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if (j == n) return i;
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}
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return -1;
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}
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/// <summary>
64+
/// assume that there are more than 10 people, and make 10th people as a celebrity.
65+
/// </summary>
66+
/// <param name="a"></param>
67+
/// <param name="b"></param>
68+
/// <returns></returns>
69+
private static bool knows(int a, int b)
70+
{
71+
if (a == b)
72+
return true;
73+
74+
if (a == 10)
75+
return false;
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return true; // false
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}
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}
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}

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