翻转后1的数量(dp)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String s = sc.next();
        
        int cnt1 = 0;  // 原本1的个数
        int[] gain = new int[n + 1];
        
        // 构造gain数组
        for (int i = 1; i <= n; i++) {
            if (s.charAt(i - 1) == '1') {
                cnt1++;
                gain[i] = -1;  // 翻转会减少一个1
            } else {
                gain[i] = 1;   // 翻转会增加一个1
            }
        }
        
        // DP数组
        int[] dp = new int[n + 1];
        dp[1] = gain[1];
        int maxGain = dp[1];
        
        // 动态规划求最大子段和
        for (int i = 2; i <= n; i++) {
            dp[i] = Math.max(gain[i], dp[i - 1] + gain[i]);
            maxGain = Math.max(maxGain, dp[i]);
        }
        
        System.out.println(cnt1 + maxGain);
    }
}

最大子段和问题