Python_Array_Problems

最近看了点视频，用代码记录下，今天是Ａrray.

1. find the most frequently occurred item in an Array

思路很简单，就是用词典来记录item出现的次数就可以，key 是item, value是item出现的次数。
注意处理有多个item同时有最高的出现次数就可以了。　

def most_frequent(given_array):
    most_count = -1
    max_item = None

    count ={}

    for item in given_array:
        if item in count:
            count[item] += 1
        else:
            count[item] = 1
    """
        if count[item] > most_count:
            max_item = item
            most_count = count[item]
    #return (max_item,most_count)
    """
    #if there is more than 1 item
    max_value = max(count.values())

    return [(i,count[i]) for i in count if count[i] == max_value]


ll=[1,2,4,5,7,1,2,2,2,2,1,1,1]
print(most_frequent(ll))

2. 

Find two common elements in 2 sorted Arrays

在两个已经排序的数组里面找到相同的element.

def common_elem(A,B):
    p1=0
    p2=0
    result = [] #List for Python and ArrayList for Java
    while p1 < len(A) and p2 < len(B):
        if A[p1] == B[p2]:
            result.append(A[p1])
            p1 += 1
            p2 += 1
        elif A[p1] > B[p2]:
            p2 += 1
        else:
            p1 += 1
    return result

l1=[1,3,4,6,7,9]
l2=[1,2,4,5,9,10]
print(common_elem(l1, l2))

3.

is One Array a rotation of Another Array

前提条件：no duplicate in Array A AND Array B

def is_rotation(A, B):
    if len(A) != len(B):
        return False

    key = A[0]
    key_i= -1

    for i in range(len(B)):
        if B[i] == key:
            key_i = i
            break
    if key_i == -1:
        return False

    for i in range(len(A)):
        j= (i+key_i) % len(A)
        if A[i] != B[j]:
            return False
    return True

la1=[1,2,3,4,5,6,7]
la2=[4,5,6,7,1,2,3]

print(is_rotation(la1, la2))