find-the-town-judge

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Algorithms

In a town, there are <code>n</code> people labeled from <code>1</code> to <code>n</code>. There is a rumor that one of these people is secretly the town judge.

If the town judge exists, then:

<ol>
	<li>The town judge trusts nobody.</li>
	<li>Everybody (except for the town judge) trusts the town judge.</li>
	<li>There is exactly one person that satisfies properties 1 and 2.</li>
</ol>

You are given an array <code>trust</code> where <code>trust[i] = [ai, bi]</code> representing that the person labeled <code>ai</code> trusts the person labeled <code>bi</code>. If a trust relationship does not exist in <code>trust</code> array, then such a trust relationship does not exist.

Return the label of the town judge if the town judge exists and can be identified, or return <code>-1</code> otherwise.

&nbsp;
Example 1:

<pre>
Input: n = 2, trust = [[1,2]]
Output: 2
</pre>

Example 2:

<pre>
Input: n = 3, trust = [[1,3],[2,3]]
Output: 3
</pre>

Example 3:

<pre>
Input: n = 3, trust = [[1,3],[2,3],[3,1]]
Output: -1
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= trust.length &lt;= 104</code></li>
	<li><code>trust[i].length == 2</code></li>
	<li>All the pairs of <code>trust</code> are unique.</li>
	<li><code>ai != bi</code></li>
	<li><code>1 &lt;= ai, bi &lt;= n</code></li>
</ul>

findJudge

Find the Town Judge

Find the Celebrity

find-the-celebrity

搜寻名人

Graph

Array

Hash Table

小镇里有 <code>n</code> 个人，按从 <code>1</code> 到 <code>n</code> 的顺序编号。传言称，这些人中有一个暗地里是小镇法官。

如果小镇法官真的存在，那么：

<ol>
	<li>小镇法官不会信任任何人。</li>
	<li>每个人（除了小镇法官）都信任这位小镇法官。</li>
	<li>只有一个人同时满足属性 1 和属性 2 。</li>
</ol>

给你一个数组 <code>trust</code> ，其中 <code>trust[i] = [ai, bi]</code> 表示编号为 <code>ai</code> 的人信任编号为 <code>bi</code> 的人。

如果小镇法官存在并且可以确定他的身份，请返回该法官的编号；否则，返回 <code>-1</code> 。

&nbsp;

示例 1：

<pre>
输入：n = 2, trust = [[1,2]]
输出：2
</pre>

示例 2：

<pre>
输入：n = 3, trust = [[1,3],[2,3]]
输出：3
</pre>

示例 3：

<pre>
输入：n = 3, trust = [[1,3],[2,3],[3,1]]
输出：-1
</pre>
&nbsp;

提示：

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= trust.length &lt;= 104</code></li>
	<li><code>trust[i].length == 2</code></li>
	<li><code>trust</code> 中的所有<code>trust[i] = [ai, bi]</code> 互不相同</li>
	<li><code>ai != bi</code></li>
	<li><code>1 &lt;= ai, bi &lt;= n</code></li>
</ul>