leetcode-1422-easy

Maximum Score After Splitting a String

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.

思路一：暴力求解

    public int maxScore(String s) {
        char[] chars = s.toCharArray();
        int ans = 0;

        for (int i = 0; i < chars.length - 1; i++) {

            int temp = 0;
            for (int j = 0; j <= i; j++) {
                if (chars[j] == '0') {
                    temp++;
                }
            }

            for (int j = i + 1; j < chars.length; j++) {
                if (chars[j] == '1') {
                    temp++;
                }
            }
            ans = Math.max(ans, temp);
        }

        return ans;
    }

思路二：先统计 1 的个数，然后从左往右遍历，对 0 的个数进行累加，然后对比两者的和，只用两次遍历就可求解

    public int maxScore(String s) {
        char[] chars = s.toCharArray();

        int oneCount = 0;
        for (char c : chars) {
            if (c == '1') {
                oneCount++;
            }
        }

        int zeroCount = 0;
        int ans = 0;
        for (int i = 0; i < chars.length - 1; i++) {
            if (chars[i] == '0') {
                zeroCount++;
            } else {
                oneCount--;
            }
            ans = Math.max(ans, zeroCount + oneCount);
        }

        return ans;
    }