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
.
More practice:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
=====================
状态转移方程为:
/**dp
f[i] = max{f[i-1]+nums[i],nums[i]},1<=i<n
return max{f[i]} 0<=i<n
*/
f[i] = max{f[i-1]+nums[i],nums[i]},1<=i<n
return max{f[i]} 0<=i<n
*/
---------
=========
code
class Solution { public: int maxSubArray(vector<int> &nums){ /**dp f[i] = max{f[i-1]+nums[i],nums[i]},1<=i<n return max{f[i]} 0<=i<n */ int length = nums.size(); vector<int> f(length); f[0] = nums[0]; for(int i = 1;i<length;i++){ f[i] = max(f[i-1]+nums[i],nums[i]); } vector<int>::iterator iter; iter = max_element(f.begin(),f.end()); return *iter; } };