# 14-line 4 ms C solution, easy understand

• Initialize 't' =0, 'max'=s[0];

Only when t > 0, 't' is a valid number, let t plus s[i]. Else reset t = s[i].

``````int maxSubArray(int* s, int ns)
{
int t = 0;
int max = s[0];
for (int i = 0; i < ns; ++i)
{
if (t>0)
t += s[i];
else
t = s[i];
max = max > t ? max : t;
}
return max;
}
``````

PS: May be the above solution is suitable for this problem (partly because this problem is not difficult), but it not a general DP solution. For example when come to #188 Best Time to Buy and Sell Stock IV. The followed DP solution is useful for this problem and #188

``````int maxSubArray(vector<int>& a)
{
if (a.empty())
return 0;
int sz = a.size();
int local; //when a[i] has been used, in this situation, the max value
int global; //final max value; global has a choice: whether add a[i] into calculation.
global = local = a[0];
int i;
for (i = 1; i < sz; ++i)
{
local = max(local + a[i], a[i]); //'a[i]' means abandon a[i-1], restart from a[i]; 'local + a[i]' means a[i-1] and a[i] all will be adopted.
global = max(global, local); //whether add a[i] into calculation.
}
return global;
}``````

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