Question

我正在尝试学习算法/数据结构。为了提高我的知识，我试图解决一些在线问题。我试图解决的问题之一是practiceque

我试过以下方法：

def count_zero_one_two():
    s = '102100211'
    s_len = len(s)
    count = 0
    for i in range (s_len-1):
        j = i+1
        k = j+1
        #print i, j, k, count
        #print s[i], s[j], s[k]
        if k > (s_len-1):
            print "end"
            break

        elif (s[i] != s[j]) and (s[i] !=s[k]) and (s[j] != s[k]):
            print s[i], s[j], s[k]
            print "not equal"
            count = count+1
            #print count
        else:
            print s[i], s[j], s[k]
            print "equal"
        k = j +i
    print count


count_zero_one_two()

问题：如果我的输入字符串是“102100211”，那么计数应为5，但我得到4.任何想法？

Answer 1

我会这样解决：

def count_zero_one_two(s):
    num = 0
    for i in range(len(s)):
        for j in range(1, len(s)/3 + 1):
            if all(s[i:i+3*j].count(n) == j for n in '012'):
                num += 1
    return num

all()用于检查所有3个字符（每次迭代）是否都在'012'。

内部for循环用于计算长度为3,6,9等的序列中的0,1和2的数量。

<强>输出：

>>> s = '0102010'
>>> count_zero_one_two(s)
2
>>> 
>>> s = '102100211'
>>> count_zero_one_two(s)
5

Answer 2

from collections import Counter

def countSub(s):
    result = []

    for i in range(3, len(s), 3):
        t = s[:i]
        c = list(Counter(t).values())
        if (c[0]==c[1]==c[2]):
            result.append((t, c[0]))

    return result


def count(s):
    result = []
    for i in range(len(s)-2):
        result.extend(countSub(s[i:]))

    return set(result)




ss = count("102100211")    
print("%s substrings found: " % len(ss), ss)

输出：

4 substrings found (not counting duplicates and empty strings): 
     {('021', 1), ('210021', 2), ('210', 1), ('102', 1)}

给定一个仅包含0,1或2的字符串，计算具有相同数量0，1和2

2 个答案: