积少成多

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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.

click to show more practice.

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
        */
---------
=========
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;
    }
};

 

 
posted on 2016-06-28 22:07  x7b5g  阅读(156)  评论(0编辑  收藏  举报