最大子序列和问题（C语言）

最大子序列和（maxSubSeqSum）

• 时间复杂度:T(N)=O(N3)

int MaxSubSeqSum(int arrays[],int length){
    int i,j,k,thisSum=0,maxSum=0;
    for(i=0;imaxSum)maxSum=thisSum;
        };
    }
    return maxSum;
}

最大子序列和改进1（maxSubSeqSum）

• 时间复杂度:T(N)=O(N2)

int MaxSubSeqSum(int arrays[],int length){
    int i,j,thisSum=0,maxSum=0;
    for(i=0;imaxSum)maxSum=thisSum;//如果本次求和大于最终结果，更新maxSum
        };
    }
    return maxSum;
}

最大子序列和（maxSubSeqSum）--分治法

算法复杂度:T(N)=O(NlgN)

int MaxSubSeqSum(int arrays[], int left, int right) {
    int sum = 0;
    if (left == right) {
        if (arrays[left] > 0)return arrays[left];
        else sum = 0;
    } else {
        int middle = (left + right) / 2;
        int leftSum = MaxSubSeqSum(arrays, left, middle);
        int rightSum = MaxSubSeqSum(arrays, middle + 1, right);
        int finalLeftSum = 0, thisLeftSum = 0;
        for (int i = left; i <=middle; i++) {
            thisLeftSum += arrays[i];
            if (thisLeftSum > finalLeftSum)finalLeftSum = thisLeftSum;
        }
        int finalRightSum = 0, thisRightSum = 0;
        for (int j = middle + 1; j < right; j++) {
            thisRightSum += arrays[j];
            if (thisRightSum > finalRightSum)finalRightSum = thisRightSum;
        }
        sum = finalLeftSum + finalRightSum;
        printf("left sum is %d,right sum is %d\n",finalLeftSum,finalRightSum);
        if (sum < leftSum)sum = leftSum;
        if (sum < rightSum)sum = rightSum;
    }
    return sum;
}

测试主函数

int main() {
    int array[] ={1,6,-5,4,2,-3,6};
    int result = MaxSubSeqSum(array,0,7);
    printf("result is %d\n", result);
}

运行结果

为了方便观察，我们将每次左右两边求得的最大子序列最大和都打印出来

F:\ClionProject\DataStruct\cmake-build-debug\DataStruct.exe
left sum is 1,right sum is 6
left sum is 0,right sum is 4
left sum is 7,right sum is 0
left sum is 2,right sum is 0
left sum is 6,right sum is 0
left sum is 0,right sum is 6
left sum is 6,right sum is 5
result is 11

Process finished with exit code 0

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