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(图论基础题) leetcode 997. Find the Town Judge

In a town, there are N people labelled from 1 to N.  There is a rumor that one of these people is secretly the town judge.

If the town judge exists, then:

  1. The town judge trusts nobody.
  2. Everybody (except for the town judge) trusts the town judge.
  3. There is exactly one person that satisfies properties 1 and 2.

You are given trust, an array of pairs trust[i] = [a, b] representing that the person labelled a trusts the person labelled b.

If the town judge exists and can be identified, return the label of the town judge.  Otherwise, return -1.

 

Example 1:

Input: N = 2, trust = [[1,2]]
Output: 2

Example 2:

Input: N = 3, trust = [[1,3],[2,3]]
Output: 3

Example 3:

Input: N = 3, trust = [[1,3],[2,3],[3,1]]
Output: -1

Example 4:

Input: N = 3, trust = [[1,2],[2,3]]
Output: -1

Example 5:

Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
Output: 3

 

Note:

  1. 1 <= N <= 1000
  2. trust.length <= 10000
  3. trust[i] are all different
  4. trust[i][0] != trust[i][1]
  5. 1 <= trust[i][0], trust[i][1] <= N

==================================================================

PS:记得几个月前在leetcode周赛上做这个题时,当时没有好好学图论,觉得这个题怎么这么难,还是但是标签是easy。。。最后用了一个很麻烦的方法写了。现在,学了图论后,发现这个题是真的很easy。。。看来,我还需要更加努力了

这个题是图论题,就是找入度为N-1和出度为0的点,trust他人就是出度,被trust就是入度。所以,最后这个题可以化为入度和出度差值为N-1的人。没找到就返回-1。

C++代码:

class Solution {
public:
    int findJudge(int N, vector<vector<int>>& trust) {
        vector<int> vec(N+1,0);
        for(auto t : trust){
            vec[t[0]]--;
            vec[t[1]]++;
        }
        for(int i = 1; i <= N; i++){
            if(vec[i] == N-1)
                return i;
        }
        return -1;
    }
};

 

另外附上几个月前写的代码:

class Solution {
public:
    int findJudge(int N, vector<vector<int>>& trust) {
        set<int> s1,s2;
        if(trust.size() == 0)
            return 1;
        int num;
        vector<vector<int> > ans;
        for(int i = 0; i < trust.size(); i++){
            s1.insert(trust[i][0]);
            s2.insert(trust[i][1]);
        }
        for(set<int>::iterator it = s2.begin(); it!=s2.end(); it++){
            if(s1.count(*it))
                s2.erase(*it);
        }
        if(s2.size() != 1)
            return -1;
        else{
            for(set<int>::iterator it = s2.begin(); it!=s2.end(); it++){
                num = *it;
            }
            for(int i = 0;i < trust.size(); i++){
                if(trust[i][1] == num){
                    ans.push_back(trust[i]);
                }
            }
            if(ans.size() == N - 1)
                return ans[0][1];
            else
                return -1;
        }
    }
};

 

posted @ 2019-05-17 19:19  PJCK  阅读(15)  评论(0编辑  收藏