918. Maximum Sum Circular Subarray

Medium

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

解題

Runtime: 59 ms, faster than 93.44%

Memory Usage: 7.5 MB, less than 81.97%

func maxSubarraySumCircular(nums []int) int {
    maxL, maxCur := nums[0], nums[0]
    minL, minCur := nums[0], nums[0]
    sum := nums[0]

 
    for i:=1; i<len(nums); i++ {
        sum += nums[i] // 記錄sum

        maxCur = max(nums[i], maxCur + nums[i]) // 記錄現在最大連續subarray值
        maxL = max(maxCur, maxL) // 記錄目前為止最大連續subarray值

        minCur = min(nums[i], minCur + nums[i]) // 記錄現在最小連續subarray值
        minL = min(minCur, minL) // 記錄目前為止最小連續subarray值
    }

    if sum == minL {
        return maxL // 一整個 array 都是負數
    }
       //最大 subarray 在中間 // 在兩邊 = 總和剪掉最小 subarray
    return max(maxL, sum - minL)
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}