【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('')
        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 == '':
                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