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.length1 <= n <= 3 * 104-3 * 104 <= nums[i] <= 3 * 104
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
}