Question

给出一个字符串，说s="##$$$#"，如何找到索引前的“＃”符号数量等于索引后的“ $”符号数量的索引？

示例：如果为s="##$$$#"，则输出为2。

说明：索引2之前有2个“＃”符号，索引2之后有2个“ $”符号

我尝试先找到中间索引并计算两侧的符号（＃和$），如果它们相等，则打印中间索引，否则增加中间数并以相同的方式进行。但是我无法正确理解逻辑。

Answer 1

一种方法是：

At any given index 'i' ,
Count_Pound[i] = Count_Pound[i-1] + 1 if charAt(i-1) is '#' 
               = Count_Pound[i-1] if charAt(i-1) is '$'
E.g. Count_Pound[0] = 0;
     Count_Pound[1] = 0 + 1 if charAt(0) = '#'
                    = 0 if charAt(0) is '$'

类似的方法适用于反向移动。

Count_Dollar[j] = Count_Dollar[j+1] + 1 if charAt(j+1) is '$'
                = Count_Dollar[j+1] if charAt(j+1) is '#'
E.g. Count_Dollar[str.length -1] = 0
     Count_Dollar[str.length - 2] = 0 + 1 if charAt(str.length-1) is '$'
                                  = 0 if charAt(str.length-1) is '#'

在遍历正向和反向方式后有了这两个数组-顺便说一句，您可以仅在一个循环中构建这两个数组，并具有两个索引，一个递增和一个递减。然后遍历这些数组并获得最大的数组（我假设您想要最大的数组）。

for i : 0 to str.length-1:
    if Count_Pound[i] == Count_Dollar[i] && max_value < Count_Pound[i]
          max_value = Count_Pound[i]
          ans = i

return ans

空间复杂度不错：O（n），时间复杂度不错：O（n）

Answer 2

您可以尝试使用for循环来测试字符串中的每个位置吗？据我所知，似乎当您说索引2是答案时，索引本身就被排除在外了，所以您得到str[:index]到str[index+1:]

所以你会得到这样的东西：

def half(str, first_half, second_half):
    for i in range(len(str)):
        if str[:i].count(first_half) == str[i+1:].count(second_half):
            return i
    return None

Answer 3

这应该缩放为O（N）：

def same_count(s):
    c1 = 0
    c2 = 0
    l1 = list()
    l2 = list()
    n = len(s)
    for i in range(n):
        # count forward from start
        if s[i] == '#':
            c1 += 1
        # count backward from end
        if s[n - i - 1] == '$':
            c2 += 1
        l1.append(c1)
        l2.append(c2)
    # reverse second list
    l2 = list(reversed(l2))
    # find index where counts are the same
    match = [i for i in range(1,n-1) if 
         l1[i-1] == l2[i+1]]
    if match:
        return match[0]
    else:
        return None


print(same_count('##$$$#'))
print(same_count('###$$$##'))
print(same_count('#$$'))

#output:
#2
#None
#1

Answer 4

string = '##$$$#'
foundHash = False
foundDollar = False
dollarIndex = -1
hashCount = 0
dollarCount = 0

for i,c in enumerate(string):
    if c == '#' and not foundHash:
        hashCount = hashCount + 1
        foundHash = True
    elif c == '#' and foundHash and not foundDollar:
        hashCount = hashCount + 1
    elif c == '$' and foundHash and not foundDollar:
        dollarIndex =  i
        foundDollar = True
    elif c == '$' and foundHash  and foundDollar:
        dollarCount =  dollarCount + 1
    else:
        if hashCount == dollarCount:
            break
        else:
            foundHash = False
            foundDollar = False
            dollarIndex = -1
            hashCount = 0 
            dollarCount = 0
            if c == '#' and not foundHash:
                hashCount = hashCount + 1
                foundHash = True

if dollarIndex > 0 and hashCount == dollarCount:
    print("index = %d" %(dollarIndex))
else:
    print("index not found")

在字符串中找到索引，其中左半部分的计数（符号）=右半部分的计数（符号）？

4 个答案: