C++動態(tài)規(guī)劃計算最大子數(shù)組

更新時間：2022年06月30日 10:35:57 作者：成就一億技術人

所謂最大子數(shù)組就是連續(xù)的若干數(shù)組元素,如果其和是最大的,那么這個子數(shù)組就稱為該數(shù)組的最大子數(shù)組

例題

題目：輸入一個整形數(shù)組，數(shù)組里有正數(shù)也有負數(shù)。數(shù)組中連續(xù)的一個或多個整數(shù)組成一個子數(shù)組，每個子數(shù)組都有一個和。求所有子數(shù)組的和的最大值。要求時間復雜度為O(n)。

例如輸入的數(shù)組為1, -2, 3, 10, -4, 7, 2, -5，和最大的子數(shù)組為3, 10, -4, 7, 2，因此輸出為該子數(shù)組的和18。

1.求最大的子數(shù)組的和

代碼【C++】

#include <iostream>
using namespace std;
/
// Find the greatest sum of all sub-arrays
// Return value: if the input is valid, return true, otherwise return false
/
bool FindGreatestSumOfSubArray
    (
    int *pData,           // an array
    unsigned int nLength, // the length of array
    int &nGreatestSum     // the greatest sum of all sub-arrays
    )
{
    // if the input is invalid, return false
    if((pData == NULL) || (nLength == 0))
        return false;
    int nCurSum = nGreatestSum = 0;
    for(unsigned int i = 0; i < nLength; ++i)
    {
        nCurSum += pData[i];
        // if the current sum is negative, discard it
        if(nCurSum < 0)
            nCurSum = 0;
        // if a greater sum is found, update the greatest sum
        if(nCurSum > nGreatestSum)
            nGreatestSum = nCurSum;
    }
    // if all data are negative, find the greatest element in the array
    if(nGreatestSum == 0)
    {
        nGreatestSum = pData[0];
        for(unsigned int i = 1; i < nLength; ++i)
        {
            if(pData[i] > nGreatestSum)
                nGreatestSum = pData[i];
        }
    }
    return true;
}
int main()
{
    int arr[] = {1, -2, 3, 10, -4, 7, 2, -5};
    int iGreatestSum;
    FindGreatestSumOfSubArray(arr, sizeof(arr)/sizeof(int), iGreatestSum);
    cout << iGreatestSum << endl;
    return 0;
}

結(jié)果

2.求和最大的相應子數(shù)組

代碼【C++】

#include <iostream>
using namespace std;
/
// Find the greatest sum of all sub-arrays
// Return value: if the input is valid, return true, otherwise return false
/
bool FindGreatestSumOfSubArray
    (
    int *pData,           // an array
    unsigned int nLength, // the length of array
    int &nGreatestSum,    // the greatest sum of all sub-arrays
    int &start,                            // Added
    int &end                            // Added
    )
{
    // if the input is invalid, return false
    if((pData == NULL) || (nLength == 0))
        return false;
    int nCurSum = nGreatestSum = 0;
    int curStart = 0, curEnd = 0;        // Added
    start = end = 0;                    // Added
    for(unsigned int i = 0; i < nLength; ++i)
    {
        nCurSum += pData[i];
        curEnd = i;                        // Added
        // if the current sum is negative, discard it
        if(nCurSum < 0)
        {
            nCurSum = 0;
            curStart = curEnd = i + 1;    // Added
        }
        // if a greater sum is found, update the greatest sum
        if(nCurSum > nGreatestSum)
        {
            nGreatestSum = nCurSum;
            start = curStart;            // Added
            end = curEnd;                // Added
        }
    }
    // if all data are negative, find the greatest element in the array
    if(nGreatestSum == 0)
    {
        nGreatestSum = pData[0];
        start = end = 0;                // Added
        for(unsigned int i = 1; i < nLength; ++i)
        {
            if(pData[i] > nGreatestSum)
            {
                nGreatestSum = pData[i];
                start = end = i;        // Added
            }
        }
    }
    return true;
}
int main()
{
    int arr[] = {1, -2, 3, 10, -4, 7, 2, -5};
    int iGreatestSum, start, end;
    FindGreatestSumOfSubArray(arr, sizeof(arr)/sizeof(int), iGreatestSum, 
        start, end);
    cout << iGreatestSum << ": ";
    for(int i = start; i <= end; i++)
    {
        cout << arr[i] << " ";
    }
    return 0;
}

結(jié)果