<p>Given an integer array <code>nums</code>, find the <span data-keyword="subarray-nonempty">subarray</span> with the largest sum, and return <em>its sum</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-2,1,-3,4,-1,2,1,-5,4]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The subarray [4,-1,2,1] has the largest sum 6.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The subarray [1] has the largest sum 1.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,4,-1,7,8]
<strong>Output:</strong> 23
<strong>Explanation:</strong> The subarray [5,4,-1,7,8] has the largest sum 23.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> If you have figured out the <code>O(n)</code> solution, try coding another solution using the <strong>divide and conquer</strong> approach, which is more subtle.</p>

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int sum=0;
        int ans=0;
        for(int i=0;i<nums.size();i++){
            if((sum+nums[i])<0)
            sum=0;
            else
            sum+=nums[i];
            ans=max(ans,sum);
        }
        if(ans==0){
        sort(nums.begin(),nums.end());
        return nums[nums.size()-1];
        }
        return ans;
    }
};
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