Python_Array_Problems_3

接上文

7.

#given a string, determine if it is comprised of all unique character:

#eg: 'abcde' -> True

# 'abcda' -> False

这个题，第一反应是将list转换成set, 比较length, 结果就很明白了。

def uni_chars(s):
    return len(s) == len(set(s))

def uni_chars_2(s):
    chars = set()

    for letter in s:
        if letter in chars:
            return False
        else:
            chars.add(letter)

    return True

print(uni_chars('abcde'))
print(uni_chars('abcda'))
print(uni_chars_2('abcde'))
print(uni_chars_2('abcda'))

8.

#compress string:

#eg: aaabbc -> a3b2c1

#eg: AAAaaa -> A3a3

我记得以前看过某个博文讲压缩的时候讲过这个压缩.

def string_compress(s):
    r = ''
    l=len(s)

    if l == 0:
        return ''
    if l == 1:
        return s+"1"

    cnt = 1
    i = 1

    while i < l:
        if s[i] == s[i-1]:
            cnt += 1
        else:
            r = r+s[i-1]+str(cnt)
            cnt = 1

        i += 1

    r = r+s[i-1]+str(cnt)

    return r

print(string_compress('AAB'))

9.

#given a string of words, reverse all the words

这种题单行脚本就足够了

def rev_words(s):
    return ' '.join(reversed(s.split()))

def rev_words_2(s):
    return ' '.join(s.split()[::-1])

def rev_words_3(s):
    words=[]
    length = len(s)
    spaces = [' ']

    i = 0
    while i < length:
        if s[i] not in spaces:
            word_start = i
            while i < length and s[i] not in spaces:
                i += 1
            words.append(s[word_start:i])

        i += 1
    return ' '.join(reversed(words))

print(rev_words('Hi John, are you ready to go?'))
print(rev_words_2('Hi John, are you ready to go?'))
print(rev_words_3('Hi John, are you ready to go?'))

10.

#Given an array of integers (positive and negative) find the largest continuous sum

p, li { white-space: pre-wrap; }

这个题我很喜欢，有点意思，因此放在最后

def large_cont_sum(arr):
    if len(arr)==0:
        return 0

    max_sum = current_sum = arr[0]

    for num in arr[1:]:
        current_sum = max(current_sum + num, num)
        max_sum = max(current_sum, max_sum)

    return max_sum


print(large_cont_sum([1,2,-1,3,4,10,10,-10,-1]))