1422. Maximum Score After Splitting a String题目虽小但是很tricky

Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring).

The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.

 

Example 1:

Input: s = "011101"
Output: 5 
Explanation: 
All possible ways of splitting s into two non-empty substrings are:
left = "0" and right = "11101", score = 1 + 4 = 5 
left = "01" and right = "1101", score = 1 + 3 = 4 
left = "011" and right = "101", score = 1 + 2 = 3 
left = "0111" and right = "01", score = 1 + 1 = 2 
left = "01110" and right = "1", score = 2 + 1 = 3

Example 2:

Input: s = "00111"
Output: 5
Explanation: When left = "00" and right = "111", we get the maximum score = 2 + 3 = 5

Example 3:

Input: s = "1111"
Output: 3

Constraints:

• 2 <= s.length <= 500
• The string s consists of characters '0' and '1' only.

先不说暴力算法，提示里有说，先计算一遍以当前位置 i 为索引的左边的1的个数，那扫描到最后，所有1的个数就知道了，将每个位置的左边1的个数保存在prefix数组中；

然后再次扫描，累计0的个数，根据先前的prefix，用1的总数减去prefix[i]就得到以当前位置为分割，右边1的个数，然后与当前位置0的个数相加就是当前分割的score.

class Solution {
public:
    int maxScore(string s) {
        vector prefix(s.size(), 0);
        int pre = 0;
        for(int i = 0; i < s.size(); i++)
        {
            if(s[i] == '1')
            {
                pre += 1;
            }
            prefix[i] = pre; //注意要更新0的位置对应的pre值
        }

        int zeroC = 0;
        int res = INT_MIN;
        for(int i = 0; i < s.size() - 1; i++)//这里一定不能扫描到最后一个，因为就没法分成左右两边了
        {
            if(s[i] == '0')
            {
                zeroC++;
            }
            res = max(res, zeroC + (pre - prefix[i]));
        }
        return res;
    }
};

注意了上面2点，就OK了，小细节啊，真得注意。

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Maximum Score After Splitting a String.

Memory Usage: 6.7 MB, less than 100.00% of C++ online submissions for Maximum Score After Splitting a String.