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最大子数组和（最大子段和）

2024-07-02 00:15:08   235人阅读

比如对于数组[1,-2,3,5,-1,2] 最大子数组和是sum[3,5,-1,2] = 9, 我们要求函数输出子数组和的最大值,并且返回子数组的左右边界(下面函数的left和right参数).

本文我们规定当数组中所有数都小于0时，返回数组中最大的数（也可以规定返回0，只要让以下代码中maxsum初始化为0即可，此时我们要注意-1 0 0 0 -2这种情形，特别是如果要求输出子数组的起始位置时，如果是面试就要和面试官问清楚）

以下代码我们在PAT 1007. Maximum Subsequence Sum测试通过,测试main函数如下

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int main()
{
    int n;
    scanf("%d", &n);
    vector<int>vec(n);
    for(int i = 0; i < n; i++)
        scanf("%d", &vec[i]);
    int left, right;
    int maxsum = maxSum1(vec, left, right);//测试时替换函数名称
    if(maxsum >= 0)
        printf("%d %d %d", maxsum, vec[left], vec[right]);
    else printf("0 %d %d", vec[0], vec[n-1]);
}

参考：编程之美2.14 求数组的子数组之和的最大值

 

算法1：最简单的就是穷举所有的子数组，然后求和，复杂度是O（n^3）

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int maxSum1(vector<int>&vec, int &left, int &right)
{
    int maxsum = INT_MIN, sum = 0;
    for(int i = 0; i < vec.size(); i++)
        for(int k = i; k < vec.size(); k++)
        {
            sum = 0;
            for(int j = i; j <= k; j++)
                sum += vec[j];
            if(sum > maxsum)
            {
                maxsum = sum;
                left = i;
                right = k;
            }
        }
    return maxsum;
}

算法2: 上面代码第三重循环做了很多的重复工作,稍稍改进如下,复杂度为O(n^2)

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int maxSum2(vector<int>&vec, int &left, int &right)
{
    int maxsum = INT_MIN, sum = 0;
    for(int i = 0; i < vec.size(); i++)
    {
        sum = 0;
        for(int k = i; k < vec.size(); k++)
        {
            sum += vec[k];
            if(sum > maxsum)
            {
                maxsum = sum;
                left = i;
                right = k;
            }
        }
    }
    return maxsum;
}

算法3: 分治法, 下面贴上编程之美的解释, 复杂度为O(nlogn)

image

image

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//求数组vec【start，end】的最大子数组和，最大子数组边界为[left，right]
int maxSum3(vector<int>&vec, const int start, const int end, int &left, int &right)
{
    if(start == end)
    {
        left = start;
        right = left;
        return vec[start];
    }
    int middle = start + ((end - start)>>1);
    int lleft, lright, rleft, rright;
    int maxLeft = maxSum3(vec, start, middle, lleft, lright);//左半部分最大和
    int maxRight = maxSum3(vec, middle+1, end, rleft, rright);//右半部分最大和
    int maxLeftBoeder = vec[middle], maxRightBorder = vec[middle+1], mleft = middle, mright = middle+1;
    int tmp = vec[middle];
    for(int i = middle-1; i >= start; i--)
    {
        tmp += vec[i];
        if(tmp > maxLeftBoeder)
        {
            maxLeftBoeder = tmp;
            mleft = i;
        }
    }
    tmp = vec[middle+1];
    for(int i = middle+2; i <= end; i++)
    {
        tmp += vec[i];
        if(tmp > maxRightBorder)
        {
            maxRightBorder = tmp;
            mright = i;
        }
    }
    int res = max(max(maxLeft, maxRight), maxLeftBoeder+maxRightBorder);
    if(res == maxLeft)
    {
        left = lleft;
        right = lright;
    }
    else if(res == maxLeftBoeder+maxRightBorder)
    {
        left = mleft;
        right = mright;
    }
    else
    {
        left = rleft;
        right = rright;
    }
    return res;
}

算法4: 动态规划, 数组为vec[]，设dp[i] 是以vec[i]结尾的子数组的最大和，对于元素vec[i+1], 它有两种选择：a、vec[i+1]接着前面的子数组构成最大和，b、vec[i+1]自己单独构成子数组。则dp[i+1] = max{dp[i]+vec[i+1],  vec[i+1]}

如果不考虑记录最大子数组的位置，于是有以下代码：                本文地址

 

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int maxSum_(vector<int>&vec)
{
    int maxsum = INT_MIN, sum = 0;
    for(int i = 0; i < vec.size(); i++)
    {
        sum = max(sum + vec[i], vec[i]);
        maxsum = max(maxsum, sum);
    }
    return maxsum;
}

 

对以上代码换个写法，并记录最大子数组的位置

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int maxSum4(vector<int>&vec, int &left, int&right)
{
    int maxsum = INT_MIN, sum = 0;
    int begin = 0;
    for(int i = 0; i < vec.size(); i++)
    {
        if(sum >= 0)
        {
            sum += vec[i];
        }
        else
        {
            sum = vec[i];
            begin = i;
        }
 
        if(maxsum < sum)
        {
            maxsum = sum;
            left = begin;
            right = i;
        }
    }
    return maxsum;
}

如果数组是循环的，该如何呢

这时分两种情形（图中红色方框表示求得的最大子数组，left、right分别是子数组的开始和结尾）：

（1）如下图最大的子数组没有跨过vec[n-1]到vec[0], 这就是每循环的情况

image

（2）如下图，最大的子数组跨过vec[n-1]到vec[0]

image

对于第二种情形，相当于从原数组中挖掉了一块(vec[right+1], …, vec[left-1]) ,那么我们只要使挖掉的和最小即可，求最小子数组和最大子数组类似，代码如下，以下代码在九度oj1572首尾相连数组的最大子数组和通过测试（测试需要，以下代码当数组全是负数时，输出0）：

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int maxSumCycle(vector<int>&vec, int &left, int&right)
{
    int maxsum = INT_MIN, curMaxSum = 0;
    int minsum = INT_MAX, curMinSum = 0;
    int sum = 0;
    int begin_max = 0, begin_min = 0;
    int minLeft, minRight;
    for(int i = 0; i < vec.size(); i++)
    {
        sum += vec[i];
        if(curMaxSum >= 0)
        {
            curMaxSum += vec[i];
        }
        else
        {
            curMaxSum = vec[i];
            begin_max = i;
        }
 
        if(maxsum < curMaxSum)
        {
            maxsum = curMaxSum;
            left = begin_max;
            right = i;
        }
        ///////////////求和最小的子数组
        if(curMinSum <= 0)
        {
            curMinSum += vec[i];
        }
        else
        {
            curMinSum = vec[i];
            begin_min = i;
        }
 
        if(minsum > curMinSum)
        {
            minsum = curMinSum;
            minLeft = begin_min;
            minRight = i;
        }
    }
    if(maxsum >= sum - minsum)
        return maxsum;
    else
    {
        left = minRight+1;
        right = minLeft-1;
        return sum - minsum;
    }
}

 

参考：面试题精解之二： 字符串、数组（1）

 

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