C++动态规划计算最大子数组
作者：成就一亿技术人
所谓最大子数组就是连续的若干数组元素,如果其和是最大的,那么这个子数组就称为该数组的最大子数组
例题
题目：输入一个整形数组，数组里有正数也有负数。数组中连续的一个或多个整数组成一个子数组，每个子数组都有一个和。求所有子数组的和的最大值。要求时间复杂度为O(n)。
例如输入的数组为1, -2, 3, 10, -4, 7, 2, -5，和最大的子数组为3, 10, -4, 7, 2，因此输出为该子数组的和18。
1.求最大的子数组的和
代码【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;
}结果
2.求和最大的相应子数组
代码【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;
}结果
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