Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] has the largest sum = 6.
解题分析：结合本周所讲的divide and conquer算法，先尝试将该问题分解为子问题，具体来说就是，所要求的subarray数组，所在的位置有三种可能（设该数组的中间下标元素为mid），第一种：在下标区间[0,mid]；第二种：在下标区间[mid+1,array.size()-1];第三种：在前两个所述区间之间，这时可知中间元素array[mid]必定在subarray中，此时可以以mid为起始点分别扫描前面两个数组找到最大子数组然后相加得到最大数组；
综上 递推式为T(n)=2T(n/2)+O(n),由主定理可知复杂度为O(nlog)；贴上代码
class Solution { public: int maxSubArray(vector<int>& nums) { return calmaxsubarr(nums, 0, nums.size()-1); } private: int div_con(vector<int>&A, int begin, int end){ if(begin == end) return A[begin]; int mid = (end-begin)/2 + begin; int left = div_con(A, begin, mid); int right = div_con(A, mid+1, end); int lefthalf = A[mid]; int tmp = A[mid]; for(int i = mid -1; i >= begin; i--){ tmp += A[i]; if(tmp > lefthalf) lefthalf = tmp; } int righthalf = A[mid+1]; tmp = A[mid+1]; for(int i = mid +2; i <= end; ++i){ tmp += A[i]; if(tmp > righthalf)righthalf = tmp; } int middle = lefthalf + righthalf; int res = left > right ? left : right; res = res > middle ? res : middle; return res; } }; 总结：该题主要难点在于递归式子的思考，查看资料得知还可以用复杂度为O(n)的dp算法解决，待有时间再去研究。
