Sword finger offer 42 Maximum sum of successive subarrays

Enter an integer array , One or more consecutive integers in an array form a subarray . Find the maximum sum of all subarrays . The required time complexity is O(n).

Example 1:

Input : nums = [-2,1,-3,4,-1,2,1,-5,4]
Output : 6
explain : Continuous subarray [4,-1,2,1] And the biggest , by 6.

Tips ：

• 1 <= arr.length <= 10^5
• -100 <= arr[i] <= 100

Code:

int maxSubArray(vector<int>& nums)
{
    
    int maxnum=nums[0];
    int sum=0;
    for(int i=0;i<nums.size();i++)
    {
    

        sum=max(sum+nums[i],nums[i]);
        maxnum=max(maxnum,sum);

    }
    cout<<maxnum<<endl;

    return maxnum;
}