2

我需要一些提示来找到返回5总和最接近或等于给定数字的元素的算法。

这些元素是大于的数字，0并且在尝试获取给定数字时应仅“使用”一次。

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！

4

2 回答 2

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]]
于 2013-01-14T01:03:44.543 回答
0

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

如果我的理解是正确的，这个问题可以通过DP解决。它与背包问题非常相似。时间复杂度为 O(n*S) 其中 n 是数组的大小，S 是您的目标

于 2013-01-14T02:26:13.390 回答