python面试题——数组

1.anagram查看两个数组是不是相同的字符组成

def anagram2(s1,s2):
    
    # Remove spaces and lowercase letters
    s1 = s1.replace(' ','').lower()
    s2 = s2.replace(' ','').lower()
    
    # Edge Case to check if same number of letters
    if len(s1) != len(s2):
        return False
    
    # Create counting dictionary (Note could use DefaultDict from Collections module)
    count = {}
    
    
        
    # Fill dictionary for first string (add counts)
    for letter in s1:
        if letter in count:
            count[letter] += 1
        else:
            count[letter] = 1
            
    # Fill dictionary for second string (subtract counts)
    for letter in s2:
        if letter in count:
            count[letter] -= 1
        else:
            count[letter] = 1
    
    # Check that all counts are 0
    for k in count:
        if count[k] != 0:
            return False

    # Otherwise they're anagrams
    return True

核心是通过字典来计算每个字母出现的字数。

2. array pair sum 

输入：pair_sum([1,3,2,2],4) 输出： (1,3) (2,2)

def pair_sum(arr,k):
    
    if len(arr)<2:
        return
    
    # Sets for tracking
    seen = set()
    output = set()
    
    # For every number in array
    for num in arr:
        
        # Set target difference
        target = k-num
        
        # Add it to set if target hasn't been seen
        if target not in seen:
            seen.add(num)
        
        else:
            # Add a tuple with the corresponding pair
            output.add( (min(num,target),  max(num,target)) )
    
    
    # FOR TESTING
    return len(output)
    # Nice one-liner for printing output
    #return '\n'.join(map(str,list(output)))

核心是通过set元素不重复的原则来避免出现重复的组合。

3.find the missing element

O(NlogN):

def finder(arr1,arr2):
    
    # Sort the arrays
    arr1.sort()
    arr2.sort()
    
    # Compare elements in the sorted arrays
    for num1, num2 in zip(arr1,arr2):
        if num1!= num2:
            return num1
    
    # Otherwise return last element
    return arr1[-1]

O(N):

import collections

def finder2(arr1, arr2): 
    
    # Using default dict to avoid key errors
    d=collections.defaultdict(int) 
    
    # Add a count for every instance in Array 1
    for num in arr2:
        d[num]+=1 
    
    # Check if num not in dictionary
    for num in arr1: 
        if d[num]==0: 
            return num 
        
        # Otherwise, subtract a count
        else: d[num]-=1 

4.Setence Reversal 把字符串反转过来

def rev_word3(s):
    """
    Manually doing the splits on the spaces.
    """
    
    words = []
    length = len(s)
    spaces = [' ']
    
    # Index Tracker
    i = 0
    
    # While index is less than length of string
    while i < length:
        
        # If element isn't a space
        if s[i] not in spaces:
            
            # The word starts at this index
            word_start = i
            
            while i < length and s[i] not in spaces:
                
                # Get index where word ends
                i += 1
            # Append that word to the list
            words.append(s[word_start:i])
        # Add to index
        i += 1
        
    # Join the reversed words
    return " ".join(reversed(words))

5.String compression 把相同的字母用个数代替

compress('AAAAABBBBCCCC') 

output:

'A5B4C4'

def compress(s):
    """
    This solution compresses without checking. Known as the RunLength Compression algorithm.
    """
    
    # Begin Run as empty string
    r = ""
    l = len(s)
    
    # Check for length 0
    if l == 0:
        return ""
    
    # Check for length 1
    if l == 1:
        return s + "1"
    
    #Intialize Values
    last = s[0]
    cnt = 1
    i = 1
    
    while i < l:
        
        # Check to see if it is the same letter
        if s[i] == s[i - 1]: 
            # Add a count if same as previous
            cnt += 1
        else:
            # Otherwise store the previous data
            r = r + s[i - 1] + str(cnt)
            cnt = 1
            
        # Add to index count to terminate while loop
        i += 1
    
    # Put everything back into run
    r = r + s[i - 1] + str(cnt)
    
    return r