LeetCode:Maximum Subarray

题目链接

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.

---

可以参考我的另一篇博文 最大子数组和（最大子段和）。

下面分别给出O(n)的动态规划解法和O(nlogn)的分治解法                              本文地址

class Solution {
public:
    int maxSubArray(int A[], int n) {
        //最大字段和问题
        int res = INT_MIN, sum = -1;
        for(int i = 0; i < n; i++)
        {
            if(sum > 0)
                sum += A[i];
            else sum = A[i];
            if(sum > res)res = sum;
        }
        return res;
    }
};

 

 

class Solution {
public:
    int maxSubArray(int A[], int n) {
        //最大字段和问题
        return helper(A, 0, n-1);
    }
private:
    int helper(int A[], const int istart, const int iend)
    {
        if(istart == iend)return A[iend];
        int middle = (istart + iend) / 2;
        int maxLeft = helper(A, istart, middle);
        int maxRight = helper(A, middle + 1, iend);
        int midLeft = A[middle];
        int tmp = midLeft;
        for(int i = middle - 1; i >= istart; i--)
        {
            tmp += A[i];
            if(midLeft < tmp)midLeft = tmp;
        }
        int midRight = A[middle + 1];
        tmp = midRight;
        for(int i = middle + 2; i <= iend; i++)
        {
            tmp += A[i];
            if(midRight < tmp)midRight = tmp;
        }
        return max(max(maxLeft, maxRight), midLeft + midRight);
    }
};

 

【版权声明】转载请注明出处：http://www.cnblogs.com/TenosDoIt/p/3713525.html