문제
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:
- The town judge trusts nobody.
- Everybody (except for the town judge) trusts the town judge.
- 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.
제한사항
- 1 <= N <= 1000
- trust.length <= 10000
- trust[i] are all different
- trust[i][0] != trust[i][1]
- 1 <= trust[i][0], trust[i][1] <= N
입출력 예
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| Example 1:
Input: N = 2, trust = [[1,2]]
Output: 2
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| Example 2:
Input: N = 3, trust = [[1,3],[2,3]]
Output: 3
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| Example 3:
Input: N = 3, trust = [[1,3],[2,3],[3,1]]
Output: -1
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| Example 4:
Input: N = 3, trust = [[1,2],[2,3]]
Output: -1
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| Example 5:
Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
Output: 3
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풀이
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| func findJudge(N int, trust [][]int) int {
item := make([][]int, N+1)
for i := range item {
item[i] = make([]int, N+1)
}
for _, row := range trust {
item[row[0]][row[1]] = 1
}
for i := 1 ; i <= N ; i++ {
for j := 1 ; j <=N ; j += 1 {
item[0][i] += item[j][i];
item[i][0] += item[i][j];
}
}
for i := 1 ; i <= N ; i++ {
if(item[0][i] == N-1 && item[i][0] == 0) {
return i;
}
}
return -1
}
|
