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.
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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[] = nums[];
for(int i = ;i<length;i++){
f[i] = max(f[i-]+nums[i],nums[i]);
}
vector<int>::iterator iter;
iter = max_element(f.begin(),f.end());
return *iter;
}
};