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.

此題不難，但翻轉數字以及判斷數字翻轉條件處很有意思。

class Solution {
public:
    int minFlipsMonoIncr(string s) {
        int len=s.size();
        int maxx=0;
        int n0=0,n1=0;
        int k=-1;
        int right=len-1,left=0;
        while(s[right]=='1')
            --right;
        while(s[left]=='0')
            ++left;
        for(int i=0;i<=right;++i)
        {
            if(s[i]=='0')
                ++n0;
            else
                ++n1;
            if(n0-n1>maxx)
            {
                maxx=n0-n1;
                k=i;
            }
        }
        if(k<=left)
        {
            n0=0;
            n1=0;
            for(int i=0;i<len;++i)
            {
                if(s[i]=='0')
                    n0++;
                else
                    n1++;
            }
            int c=len-1;
            int d=0;
            while(s[c]=='1')
            {
                --c;
                ++d;
            }
            return min(n0-left,n1-d);
        }
        else
        {
            int count=0;
            for(int i=0;i<=k;++i)
                if(s[i]=='1')
                    count++;
            for(int i=k+1;i<len;++i)
                if(s[i]=='0')
                    count++;
            return count;
        }
    }
};