twitter online 數組中找集合的問題

此處已經給出了三種解法：勇幸|Thinking 點擊開啟連結

 

但是我還是想記錄一下。

 

題目：

A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range [0..N−1]. Sets S[K] for 0 ≤ K < N are defined as follows:S[K]
= { A[K], A[A[K]], A[A[A[K]]], ... }. Sets S[K] are finite for each K.

Write a function:

class Solution { public int solution(int[] A); }

that, given an array A consisting of N integers, returns the size of the largest set S[K] for this array. The function should return 0 if the array is empty.

 

注意：

1） 數組為空白，元素為負數

2）A[A[i]]越界

3）環路的發現和終止

4）數組中的所有數字各不相同

執行個體： a =[1,2,3,9,6,5,8,7,4,11,-1]

環路a[4]=6,a[6]=8,a[8]=4; a[5]= 5；a[7]=7

 

暴力解法：

a = [1,2,3,9,6,5,8,7,4,11,-1]flag = []n = len(a)def bfcounter(n):    maxv = 0    j = 0       for k in xrange(n):        count = 0        j = k           while j <n and j >=0 and a[j]!=j and  flag[j]==False:            count+=1            flag[j]=True            j =a[j]                 if j<n and j >=0  and a[j]==j:            count+=1                if count >maxv:            maxv = count     return maxvfor k in xrange(n):    flag.append(False)maxv = bfcounter(n)print "max  is ",maxv

使用flag標記是否已經被計算過，防止陷入環路中。

 

另外一種解法是遞迴的方法，減少重複計算。總的是基於 count(i)=1+count(a[i])，遞迴計算的同時也要防止陷入環路中。

python 代碼：

 

#!/usr/bin/pythonimport osimport sysa = [1,2,3,9,6,5,8,7,4,11，-1]n =len(a) flag = []def countSet(i,M,n):    count =1     j=a[i]      flag[i]=True    if  j >=n or j <0  or j ==i:        M[i]=1          return count    if j < n and j >=0 :        if M[j]==0 and flag[j]==False:# not counted and not being computed.            count = 1+countSet(j,M,n)                elif M[j]==0 and flag[j]:            # runs into a cycle,then stop            count =1            M[j]=1                      return count        else:               return count                    count = 1+M[j]        M[i]=count        return count    return countdef main():     M = []      for k in xrange(n):        M.append(0)        flag.append(False)            maxv = 0    for k in xrange(n):        if M[k]==0:            tmp = countSet(k,M,n)            if tmp >maxv:                maxv = tmp            M[k]=tmp    print "M:",M    print "max set size is ",maxvif __name__=="__main__":    main()  

注意：計算完成後，M中某些對應項的統計量不一定是最終的結果。例如對4，6，8的這個環而言，M[8]=1,M[6]=2,M[4]=3 。這是由於flag的作用。。。。（我說的是不是太白癡了……）

  but！反正不影響結果，我們只要最大的嘛~

 

 

並查集的解法：

雖然不是第一次看到並查集的講解，但是卻是第一次抄寫相關的代碼……

python 代碼

 
a = [1,2,3,9,6,5,8,7,4,11,-1]parent =[]n=len(a)num =[]maxv = 0def findset(x):#x is index??    while (x >=0 and x<n) and x != parent[x]:        x=parent[x]            if x <0 or x>=n:        return -1    return parent[x]def unionset(i,j):#i is index ,j = a[i]??    global maxv       x = findset(i)    if x <0:        return         if j <0 or j >=n:        num[x]+=1        return          y = findset(j)    if x ==y:        return          parent[x]=y    num[y]+=num[x]    num[x]=num[y]    if maxv< num[x]:         maxv=num[x]for i in xrange(n):    num.append(1)    parent.append(i)for i in xrange(n):    unionset(i,a[i])print "num ",numprint "maxv ",maxv