经典算法——求最大子序列和(2)

这是我面试的时候想到的算法的实现，使用分治法，算法复杂度为O(n*log(n))。算法描述如下：
     对于每一个划分子序列需要获取4个数值：
         sum（子序列和）、maxSum（最大子序列和）、lMaxSum（最大的含有最左侧节点的子序列和）、rMaxSum（最大的含有最右侧节点的子序列和）
     递归算法（res为需要运算的结果，lRes、rRes分别为该段的左右划分）：
         res->sum = lRes->sum + rRes->sum;
         res->lMaxSum = max(lRes->lMaxSum, lRes->sum + rRes->lMaxSum);
         res->rMaxSum = max(rRes->rMaxSum, rRes->sum + lRes->rMaxSum);
         res->maxSum = max3(lRes->maxSum, rRes->maxSum, lRes->rMaxSum + rRes->lMaxSum);

全部源代码如下：

/**//*
 * 子序列结果 
 */
typedef struct _SubSequenceRes 
...{
    int sum;
    int maxSum;
    int lMaxSum;
    int rMaxSum;
} SubSequenceRes;

/**//*
 * 求两数中的最大值 
 */
static int max(int m, int n)
...{
    return (m>n) ? m : n;
}

/**//*
 * 求三数中的最大值 
 */
static int max3(int a, int b, int c)
...{
    return max(a, max(b, c));
}

/**//*
 * 创建子序列结果结构 
 */
static SubSequenceRes * newSubSequenceRes()
...{
    return (SubSequenceRes *)malloc(sizeof(SubSequenceRes));
}

/**//*
 * 释放子序列结果结构 
 */
static void deleteSubSequenceRes(SubSequenceRes *pointer)
...{
    free(pointer);
}

/**//*
 * 循环解析子序列，获得子序列结果结构
 */
static SubSequenceRes * _maxSubSequenceSum4(int a[], int left, int right)
...{
    int center;
    int leftMaxSum;
    int rightMaxSum;
    SubSequenceRes *res = newSubSequenceRes();
    SubSequenceRes *lRes, *rRes;
    
    if (left == right)
    ...{
        res->sum = res->maxSum = res->lMaxSum = res->rMaxSum = a[left];
    }
    else
    ...{
        center = (left + right) >> 1;    /**//* center = (left + right) / 2; */
        lRes = _maxSubSequenceSum4(a, left, center);
        rRes = _maxSubSequenceSum4(a, center + 1, right);
        res->sum = lRes->sum + rRes->sum;
        res->lMaxSum = max(lRes->lMaxSum, lRes->sum + rRes->lMaxSum);
        res->rMaxSum = max(rRes->rMaxSum, rRes->sum + lRes->rMaxSum);
        res->maxSum = max3(lRes->maxSum, rRes->maxSum, lRes->rMaxSum + rRes->lMaxSum);
        deleteSubSequenceRes(lRes);
        deleteSubSequenceRes(rRes);
    }
    
    return res;
}

/**//*
 * 分治法实现：
 *     对于每一个划分子序列需要获取4个数值：
 *         sum（子序列和）、maxSum（最大子序列和）、lMaxSum（最大的含有最左侧节点的子序列和）、rMaxSum（最大的含有最右侧节点的子序列和） 
 *     递归算法（res为需要运算的结果，lRes、rRes分别为该段的左右划分）： 
 *         res->sum = lRes->sum + rRes->sum;
 *         res->lMaxSum = max(lRes->lMaxSum, lRes->sum + rRes->lMaxSum);
 *         res->rMaxSum = max(rRes->rMaxSum, rRes->sum + lRes->rMaxSum);
 *         res->maxSum = max3(lRes->maxSum, rRes->maxSum, lRes->rMaxSum + rRes->lMaxSum);
 */
int maxSubSequenceSum4(int a[], int len)
...{
    SubSequenceRes * res = _maxSubSequenceSum4(a, 0, len - 1);
    int result = res->maxSum;
    deleteSubSequenceRes(res);
    return result;
}