Flip String to Monotone Increasing

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.

题目理解：

给定一个只包含0和1的数组，一个flip表示将一个0换成1或者将1换成0，问最少经过多少次flip，才能将数组变为前面都是0，后面都是1的形式

解题思路：

记录从每一个位置为划分，将其前面变为0，后面变为1的filp数目，返回最小的。这里的技巧是，从前往后遍历一遍，记录从当前位置之前有多少个1，从后往前遍历一遍，记录当前位置之后有多少个0，将同一个位置的两个统计量相加，就是filp次数

代码如下：

class Solution {
    public int minFlipsMonoIncr(String S) {
        int len = S.length();
        int[] a = new int[len], b = new int[len];
        char[] chs = S.toCharArray();
        for(int i = 0 ; i < len; i++){
            if(i > 0)
                a[i] = a[i - 1];
            if(chs[i] == '1')
                a[i]++;
        }
        for(int i = len - 1; i > -1; i--){
            if(i < len - 1)
                b[i] = b[i + 1];
            if(chs[i] == '0')
                b[i]++;
        }
        int res = len;
        for(int i = 0; i < len; i++){
            if(i == 0)
                res = Math.min(res, b[1]);
            else if(i == len - 1)
                res = Math.min(res, a[len - 1]);
            else
                res = Math.min(res, a[i - 1] + b[i + 1]);
        }
        return res;
    }
}

标签：int,Monotone,Flip,len,flip,++,res,Increasing,Math
来源： https://blog.csdn.net/m0_37889928/article/details/86652444