【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

 

posted @ 2018-10-22 10:02  seyjs  阅读(456)  评论(0编辑  收藏  举报