在数组中找到5个元素，它们的和最接近或等于给定的数字

Question

我需要一些提示来找到一种算法，该算法返回5元素的总和最接近或等于给定数字。

这些元素是大于0数字，并且在尝试获取给定数字时，每个元素只能“使用” 1次。

假设我们有一个数组{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}和我们想要得到的数字21 。 它应该返回{2, 3, 4, 5, 7} 。

任何帮助是极大的赞赏！

Answer 1

ok n loops all begin with counts = 0 (in this case n = 5 )
all loops end at MainArraySize (in this case MainArraySize = 10)

int RequiredSum = ValueOfRequiredSum;
int CurrentSum = 0;
int CurrentClosestSum = 0;
int[] Finalindexesrequired = int[5]{0,0,0,0,0};

//U might want to add Duplicates 
duplicate_count ++;
int[5][] FinalindexesRequiredDuplicates= new int[5][];


for(int loopcount1 = 0, loopcount1++, loopcount < MainArraySize-1)
{
for(int loopcount2 = 0, loopcount2++, loopcount < MainArraySize-1)
{..
..
..
for(int loopcount5 = 0, loopcount5++, loopcount < MainArraySize-1)
{

-------------------------------------
Now this is all inside the 5th loop
//Process logic here
//Looping for first time
if(CurrentSum = 0)
{Currentsum = MainArray[loopcount1] + MainArray[loopcount2] + .... + MainArray[loopcount5]
CurrentClosestSum = CurrentSum
FinalindexesRequired[0] = loopcount1;
FinalindexesRequired[1] = loopcount2;
..
..
FinalindexesRequired[4] = loopcount2;
}

Currentsum = MainArray[loopcount1] + MainArray[loopcount2] + .... + MainArray[loopcount5]


if((RequiredSum - CurrentSum) < (RequiredSum - CurrentClosestSum))
{
//Am gonna change the indexes because the currentsum ITERATION is closer
FinalindexesRequired[0] = loopcount1;
FinalindexesRequired[1] = loopcount2;
..
..
FinalindexesRequired[4] = loopcount2;

//If u wanted the duplicates also, since u came to a fresher ITERATION
Reset the duplicatecount to 0 and remove all duplicates because they aint valid anymore
}

//What u Might want to Add is this
if((Requiredsum - CurrentSum) = (RequiredSum - CurrentCosestSum))
{
//Hey we got Duplicates
duplicate_count ++;
FinalindexesRequiredDuplicates[0][duplicate_count] = loopcount1;
FinalindexesRequiredDuplicates[1][duplicate_count] = loopcount1;
..
..
FinalindexesRequiredDuplicates[5][duplicate_count] = loopcount1;
}

}



-------------------------------- End of 5th loop
}}}}}



//FINALLY AFTER EXITING I THINK U HAVE UR ANSWER IN

MAINARRAY[FINALINDEXESREQUIRED[0]]
MAINARRAY[FINALINDEXESREQUIRED[1]]
MAINARRAY[FINALINDEXESREQUIRED[2]]
MAINARRAY[FINALINDEXESREQUIRED[3]]
MAINARRAY[FINALINDEXESREQUIRED[4]]


//IN CASE OF DUPLICATES U HAVE UR ANSWER IN
set 1:
MAINARRAY[FINALINDEXESREQUIRED[0]]
MAINARRAY[FINALINDEXESREQUIRED[1]]
MAINARRAY[FINALINDEXESREQUIRED[2]]
MAINARRAY[FINALINDEXESREQUIRED[3]]
MAINARRAY[FINALINDEXESREQUIRED[4]]

other sets:
MAINARRAY[FinalindexesRequiredDuplicates[0][0...n]]
MAINARRAY[FinalindexesRequiredDuplicates[1][0...n]]
MAINARRAY[FinalindexesRequiredDuplicates[2][0...n]]
MAINARRAY[FinalindexesRequiredDuplicates[3][0...n]]
MAINARRAY[FinalindexesRequiredDuplicates[4][0...n]]

Answer 2

如果目标是21，为什么不返回{2,3,4,5,7}而不是{2,3,4,5,6}？ 非常混淆...

如果我的理解是正确的，则DP可以解决此问题。 这非常类似于敲打麻袋的问题。 时间复杂度为O（n * S），其中n是数组的大小，S是您的目标