Given a circular integer arraynums of length n, return the maximum possible sum of a non-empty subarray ofnums.
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.
Input: nums = [1,-2,3,-2] Output: 3 Explanation: Subarray [3] has maximum sum 3.
Input: nums = [5,-3,5] Output: 10 Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.
Input: nums = [-3,-2,-3] Output: -2 Explanation: Subarray [-2] has maximum sum -2.
n == nums.length1 <= n <= 3 * 104-3 * 104 <= nums[i] <= 3 * 104
implSolution{pubfnmax_subarray_sum_circular(nums:Vec<i32>) -> i32{let sum = nums.iter().sum();letmut max_sum = nums[0];letmut min_sum = nums[0];letmut prefix_sum = nums[0];letmut ret = nums[0].max(sum);for i in1..nums.len(){ prefix_sum += nums[i]; ret = ret .max(prefix_sum - min_sum).max(sum - prefix_sum + max_sum); max_sum = max_sum.max(prefix_sum); min_sum = min_sum.min(prefix_sum);} ret }}