採用分治演算法，

陣列區間[left, right]， mid = (right + left) / 2

最大子陣列存在於可能的三種情況：

[left, mid],  [mid + 1, right], 橫跨前兩個

而第三種情況可以拆成[lb, mid] 和 [mid + 1, ub]的並集

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        if (nums.size() == 1) return nums[0];
        if (nums.size() == 2
 
) return max(nums[0], nums[1], nums[0] + nums[1]);

        int left = 0, right = nums.size() - 1, mid = left + (right - left) / 2;
        int left_max = nums[mid], right_max = 0, cur_left = 0, cur_right = 0;
        vector<int>left_nums, right_nums;
        for (int i = mid; i >= 0; --i)
        {
            cur_left 
 
+= nums[i];
            if (cur_left > left_max) left_max = cur_left;
            left_nums.push_back(nums[i]);
        }
        if (mid + 1 <= right) right_max = nums[mid + 1];
        for (int i = mid + 1; i <= right; ++i)
        {
            cur_right += nums[i];
            if (cur_right > right_max) right_max = cur_right;
            right_nums.push_back(nums[i]);
        }
        
 
return max(maxSubArray(left_nums), maxSubArray(right_nums), left_max + right_max);
    }

    int max(int a, int b, int c)
    {
        int m = a;
        if (m < b) m = b;
        if (m < c) m = c;
        return m; 
    }
};