题解 | #环形数组的连续子数组最大和#

感觉题解都说复杂了，琢磨了很久才搞明白，所以给大家说一下我的理解：
首先判断
1：包含最右边那个数的子序列能否大于0
2：包含最左边那个数的子序列能否大于0
如果满足那说明，最大序列有可能头尾相接
然后简单判断一下
1：普通的最大值
2：所有数之和 - 普通的最小值
这两个哪个大就输出哪个就行了
import java.io.*;
import java.util.*;

public class Main {
	public static void main(String[] args) throws IOException {
		int n = nextInt();
		int[] list = new int[n];
		int sum = 0;
		for (int i = 0; i < n; i++) {
			list[i] = nextInt();
			sum += list[i];
		}
		int dp_left = Integer.MIN_VALUE/2;
		int dp_right = Integer.MIN_VALUE/2;
		int dp_min = Integer.MAX_VALUE/2;
		int min = Integer.MAX_VALUE/2;
		int max = Integer.MIN_VALUE/2;
		for (int i = 0; i < n; i++) {
			dp_left = Math.max(list[i], dp_left + list[i]);
			max = Math.max(dp_left, max);
		}
		for (int i = n-1; i >= 0; i--)
			dp_right = Math.max(list[i], dp_right + list[i]);
		for (int i = 0; i < n; i++) {
			dp_min = Math.min(list[i], dp_min + list[i]);
			min = Math.min(dp_min, min);
		}
		
		int ans = 0;
		if(dp_left>0 && dp_right>0) 
			ans = Math.max(sum - min, max);
		else 
			ans = max;
		
		System.out.println(ans);
		
	}
	
	
	
	
	
	static PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StreamTokenizer st = new StreamTokenizer(br);
	static int nextInt() throws IOException {
		st.nextToken();
		return (int)st.nval;
	}
	static long nextLong() throws IOException {
		st.nextToken();
		return (long)st.nval;
	}
}