【LeetCode 918】 Maximum Sum Circular Subarray

題目描述

Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C.

Here, a circular array means the end of the array connects to the beginning of the array. (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)

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

Example 1:

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

Example 2:

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

Example 3:

Input: [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4

Example 4:

Input: [3,-2,2,-3]
Output: 3
Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3

Example 5:

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

Note:

-30000 <= A[i] <= 30000
1 <= A.length <= 30000

思路

第一種情況，連續子數組和。
第二種情況，首尾兩個連續子數組之和，那麼中間一段連續子數組和是最小值。

代碼

class Solution {
public:
    int maxSubarraySumCircular(vector<int>& A) {
        int cur = A[0], max_ = A[0];
        int n = A.size();
        for (int i=1; i<n; ++i) {
            cur = max(cur+A[i], A[i]);
            max_ = max(max_, cur);
        }
        
        cur = A[0];
        int min_ = A[0];
        int sum = A[0];
        for (int i=1; i<n; ++i) {
            sum += A[i];
            cur = min(cur+A[i], A[i]);
            min_ = min(min_, cur);
        }
        
        if (max_ < 0) return max_;
        return max(max_, sum - min_);
    }
};