描述
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.
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 <= N <= 1000
0 <= trust.length <= 10^4
trust[i].length == 2
trust[i] are all different
trust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N
解析
根据题意，要找出隐藏在城镇中的法官，法官必须满足其他人都相信他，他不相信任何人。现在给出了 trust 列表表示 a 相信 b ，这样就可以对每个人进行计数。用长度为 N+1 的列表 d 表示所有人（索引为 0 的元素无用，其他索引代表城镇的每个人），然后遍历 trust 中的每个子列表 [a,b] ，d[a]-=1 表示他如果去相信其他人表示他肯定不是法官， d[b]+=1 表示他被人相信则可能是法官，遍历结束之后，只要 d 中的位置的计数为 N-1 ，就表示其为法官，将其索引返回，否则返回 -1 。
解答
class Solution(object):
def findJudge(self, N, trust):
"""
:type N: int
:type trust: List[List[int]]
:rtype: int
"""
d = [0] * (N+1)
for a,b in trust:
d[a] -= 1
d[b] += 1
for person in range(1, N + 1):
if d[person] == N - 1:
return person
return -1
运行结果
Runtime: 596 ms, faster than 96.12% of Python online submissions for Find the Town Judge.
Memory Usage: 18.8 MB, less than 22.99% of Python online submissions for Find the Town Judge.
解析
另外在换一种思路，那就是我们维持两个长度为 N 列表 L 和 R （索引为 0 的元素无效），L[i] 表示第 i 个人得到的票数， R[i] 表示第 i 个人投出去多少票，法官肯定是一个得到 N-1 票，但是投出去 0 票的人，所以如果 N-1 在 L 中，那么找出其索引 idx ，然后再判断其对应的 R[idx] 是否为 0 ，如果是 0 那么表示 i 就是法官，否则直接返回 -1 即可。
解答
class Solution(object):
def findJudge(self, N, trust):
"""
:type N: int
:type trust: List[List[int]]
:rtype: int
"""
L = [0] * (N+1)
for a, b in trust:
L[b] += 1
R = [0] * (N+1)
for a, b in trust:
R[a] += 1
if N-1 in L:
idx = L.index(N - 1, 1)
if R[idx] == 0:
return idx
return -1
运行结果
Runtime: 751 ms, faster than 21.26% of Python online submissions for Find the Town Judge.
Memory Usage: 18.8 MB, less than 29.90% of Python online submissions for Find the Town Judge.
原题链接：leetcode.com/problems/fi…
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