Leetcode 1702. Maximum Binary String After Change

文章作者：Tyan
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1. Description

Maximum Binary String After Change

2. Solution

解析：Version 1，先找到第一个0，然后找到其后的第一个1，从这里开始，每碰到一个0就将其置为1，1之后的数字对应的置为0，相当于互换二者位置，这样让所有的0集中在一起，然后执行第一条规则。Version 2，根据规则可知，第一个0后面的零都应该跟其相连，即其后的数字顺序应按照0和1的顺序排序，然后将执行第一条规则。Version 3根据规则可知，如果字符串中的0少于两个，则字符串没变化，0多于1个时，最终结果只有1个0，且其位置应该位于第一个0之后的第count位，count为字符串中0的总数。

• Version 1

class Solution:
    def maximumBinaryString(self, binary: str) -> str:
        digits = list(binary)
        n = len(digits)
        i = 0
        while i < n and digits[i] != '0':
            i += 1
        m = i
        while i < n and digits[i] != '1':
            i += 1
        k = i
        for j in range(i, n):
            if digits[j] == '0':
                digits[j] = '1'
                digits[k] = '0'
                k += 1
        i = m
        while i < n - 1 and digits[i] == '0' and digits[i+1] == '0':
            digits[i] = '1'
            i += 1
        ans = ''.join(digits)
        return ans

• Version 2

class Solution:
    def maximumBinaryString(self, binary: str) -> str:
        digits = list(binary)
        n = len(digits)
        i = 0
        while i < n and digits[i] != '0':
            i += 1
        digits = digits[:i] + sorted(digits[i:])
        i = 0 
        while i < n - 1:
            if digits[i] == '0' and digits[i+1] == '0':
                digits[i] = '1'
            i += 1

        ans = ''.join(digits)
        return ans

• Version 3

class Solution:
    def maximumBinaryString(self, binary: str) -> str:
        digits = ['1'] * len(binary)
        count = binary.count('0')
        if count <= 1:
            return binary
        n = len(digits)
        i = 0
        while i < n and binary[i] != '0':
            i += 1
        digits[i+count-1] = '0'
        ans = ''.join(digits)
        return ans

Reference

1. https://leetcode.com/problems/maximum-binary-string-after-change/