题解 | #环形数组的连续子数组最大和#

先找出和最小的连续子数组，将该子数组移动到数组末尾，然后求最大连续子数组和

import java.util.Scanner;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] nums = new int[n];
        for (int i = 0; i < nums.length; i++) {
            nums[i] = in.nextInt();
        }

        System.out.println(maxSum(nums));
    }

    public static int maxSum(int[] nums) {
        int n, max, min, minIndex;
        n = nums.length;
        int[] dp = new int[n];
        minIndex = 0;
        min = dp[0] = nums[0];
        for (int i = 1; i < n; i++) {
            // dp[i] = Math.max(dp[i-1] + nums[i], nums[i]);
            // max = Math.max(max, dp[i]);

            dp[i] = Math.min(dp[i - 1] + nums[i], nums[i]);
            if (dp[i] < min) {
                min = dp[i];
                minIndex = i;
            }
        }
        //System.out.println("dp1 = " + Arrays.toString(dp));

        int[] nums2 = new int[nums.length]; // 借用一个辅助数组，对原数组做处理
        for (int i = minIndex + 1; i < n; i++) {
            nums2[i - minIndex - 1] = nums[i];
        }
        for (int i = 0; i < minIndex + 1; i++) {
            nums2[n - 1 - minIndex + i] = nums[i];
        }

        max = dp[0] = nums2[0];
        for (int i = 1; i < n; i++) {
            dp[i] = Math.max(dp[i - 1] + nums2[i], nums2[i]);
            max = Math.max(max, dp[i]);
        }
        return max;
    }
}