算法4：求整型数组的连续子数组的最大和

输入一个整型数组，数组里既有正数也有负数，数组中连续一个或者多个组成整型子数组，每个子数组都有一个和，求所有子数组的最大和，要求时间复杂度为O(n)。

解题思路

使用贪婪算法，数组从开始往后遍历，记录最大和，当前和小于0时，当前子数组和置为0.

c++代码如下

//使用贪婪算法，遇到sum<0的情况，就将sum置0void GetSubMaxSum(int srcBuffer[],int srcBufferLen,int maxSubBuffer[],int& maxSubLength,int& maxSum){    if(srcBuffer == NULL || srcBufferLen <= 0)    {        maxSubBuffer = NULL;        maxSum = 0;        return;    }    maxSum = 0;    int maxSubBufferLen = 0;    int sum = 0;    int beginIndex = 0;    int endIndex = 0;    for(int i = 0; i < srcBufferLen; i++)    {        sum +=srcBuffer[i];        if(sum > maxSum)        {            endIndex = i;            maxSum = sum;        }        if(sum <= 0)        {            beginIndex = i;            sum=0;        }    }    maxSubLength = 0;    for(int i = beginIndex; i <=endIndex; i++)    {        maxSubBuffer[maxSubLength++]=srcBuffer[i];    }}

测试代码如下

int _tmain(int argc, _TCHAR* argv[]){    const int srcLength = 10;    int srcBuffer[srcLength] = {0};    cout<<"please input 10 int valude"<<endl;    for(int i = 0; i < srcLength;i++)    {        cin>>srcBuffer[i];    }    int maxSubBuffer[srcLength] = {0};    int maxSum = 0;    int maxSubLength = 0;    GetSubMaxSum(srcBuffer,srcLength,maxSubBuffer,maxSubLength,maxSum);    cout<<"maxsum = "<<maxSum<<endl;    for(int i = 0; i < maxSubLength; i++)    {        cout<<maxSubBuffer[i]<<" ";    }    cout<<endl;    return 0;}