918. Maximum Sum Circular Subarray

https://leetcode.com/problems/maximum-sum-circular-subarray/

Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums.

A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].

A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.

Example 1:

Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3.

Example 2:

Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.

Example 3:

Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.

Constraints:

• n == nums.length

• 1 <= n <= 3 * 104

• -3 * 104 <= nums[i] <= 3 * 104

Solution:

https://www.jiuzhang.com/problem/maximum-sum-circular-subarray/

What is an alternate way of representing a circular array so that it appears to be a straight array? Essentially, there are two cases of this problem that we need to take care of. Let's look at the figure below to understand those two cases:

Case 1: #### case 1: max appear in the middle

Answer is in (LeetCode 53 Maximum Subarray)

Case 2: #### case 2: max appear at two ends, so in the middle is min ,

Find MaxSubArray, sumArray- MinSubArray (注意这里要判断sumArray != MinSubArray)

class Solution {
    public int maxSubarraySumCircular(int[] nums) {
    
        // case 1:
        int case1 = maxSubArray(nums);
        //case 2:
        int case2 = sumArray(nums) - minSubArray(nums);
        if(case2 != 0)
        {
            return Math.max(case1, case2);
        }
        
        return case1;
        
    }
    
    public int sumArray(int[] nums)
    {
        int sum = 0;
        for(int i = 0; i< nums.length; i++)
        {
            sum+=nums[i];
        }
        return sum;
    }
    
    public int maxSubArray(int[] nums)
    {
        int n = nums.length;
        int[] dp = new int[n+1];
        dp[0] = Integer.MIN_VALUE;
        for(int i=1; i<= n; i++)
        {
            if(dp[i-1] < 0)
            {
                dp[i] = nums[i-1];
            }
            else
            {
                dp[i] = dp[i-1]+nums[i-1];
            }
        }
        int result = Integer.MIN_VALUE;
        for(int i = 0; i<=n ; i++)
        {
            result = Math.max(result, dp[i]);
        }
        return result;
    }
    
    public int minSubArray(int[] nums)
    {
        int n = nums.length;
        int[] dp = new int[n+1];
        dp[0] = Integer.MAX_VALUE;
        for(int i = 1; i<= n; i++)
        {
            if(dp[i-1] > 0)
            {
                dp[i] = nums[i-1];
            }
            else
            {
                dp[i] = dp[i-1]+nums[i-1];
            }
        }
        int result = Integer.MAX_VALUE;
        for(int i = 0; i<=n ; i++)
        {
            result = Math.min(result, dp[i]);
        }
        return result;
    }
}