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  • 【leetcode】926.Flip String to Monotone Increasing

    题目如下:

    A string of '0's and '1's is monotone increasing if it consists of some number of '0's (possibly 0), followed by some number of '1's (also possibly 0.)

    We are given a string S of '0's and '1's, and we may flip any '0' to a '1' or a '1' to a '0'.

    Return the minimum number of flips to make S monotone increasing.

    Example 1:

    Input: "00110"
    Output: 1
    Explanation: We flip the last digit to get 00111.
    

    Example 2:

    Input: "010110"
    Output: 2
    Explanation: We flip to get 011111, or alternatively 000111.
    

    Example 3:

    Input: "00011000"
    Output: 2
    Explanation: We flip to get 00000000.
    

    Note:

    1. 1 <= S.length <= 20000
    2. S only consists of '0' and '1' characters.

    解题思路:转换后的字符串有两种形式,一个是全为0,另一个是前半部分全是0后半部分全是1。第一种情况的翻转次数是S中1的个数;第二种的情况,我们只要找出转换后的字符串第一个1所在的位置即可。设第一个所在的位置是i,那么S[0:i-1]区间内所有1都要变成0,而S[i:-1]区间内所有的0要变成1,总的翻转次数就是前一段区间内1的个数加上后一段区间内0的个数即可。遍历S,即可求出第二种情况的最小值,再与第一种情况求得的值比较取最小值即可。

    代码如下:

    class Solution(object):
        def minFlipsMonoIncr(self, S):
            """
            :type S: str
            :rtype: int
            """
            t_count_1 = S.count('1')
            t_count_0 = len(S) - t_count_1
            res = t_count_1
    
            count_0 = 0
            count_1 = 0
            for i,v in enumerate(S):
                if v == '0':
                    count_0 += 1
                else:
                    #count_1 += 1
                    # if convert this 1 to 0
                    res = min(res,count_1 + (t_count_0 - count_0))
                    count_1 += 1
            return res
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  • 原文地址:https://www.cnblogs.com/seyjs/p/9828593.html
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