我有一个线性时间最大连续子序列求和算法,它假设最小子序列长度仅为 0:
int maxSubSum4(const vector<int> & a, const int &minSeq)
{
int maxSum = 0, thisSum = 0;
for (int j = 0; j < a.size(); j++)
{
thisSum += a[j];
if (thisSum > maxSum)
maxSum = thisSum;
else if (thisSum < 0)
thisSum = 0;
}
return maxSum;
}
关于如何更新它以处理正用户指定的最小子序列长度(
minSeq
)的任何提示?我完全被难住了。 最佳答案
创建一个 minSeq
长度总和数组,这将是您的下限:
int maxSubSum5(const vector<int> & a, const int &minSeq){
if(minSeq > a.size()) throw logic_error("maxSubSum5 - minSeq too big!");
if(minSeq == 0) return maxSubSum4(a, minSeq);
vector<int> minimal(a.size()-minSeq+1);
minimal[0] = 0;
for(size_t i=0; i<minSeq; ++i) minimal[0] += a[i];
for(size_t i=1; i<minimal.size(); ++i) {
minimal[i] = minimal[i-1] - a[i-1] + a[i+minSeq-1];
}
int maxSum = minimal[0], currentSum = maxSum;
for(size_t i=minSeq; i<a.size(); ++i){
currentSum += a[i];
if(currentSum < minimal[i-minSeq+1]) currentSum = minimal[i-minSeq+1];
if(currentSum > maxSum) maxSum = currentSum;
}
return maxSum;
}
(每当我们重置
currentSum
时,我们都会切断子序列 s
,其中包含最后一个元素的 s
的任何子序列都具有负和。)更新: 由于我们只使用
minimal
的每个值一次,它们可以“即时”计算,而无需使用 O(N) 空间。这也使代码更短:int maxSubSum5(const vector<int> & a, const int &minSeq){
if(minSeq > a.size()) throw logic_error("maxSubSum5 - minSeq too big!");
if(minSeq == 0) return maxSubSum4(a, minSeq);
int minimalSum = 0;
for(size_t i=0; i<minSeq; ++i) minimalSum += a[i];
int maxSum = minimalSum, currentSum = minimalSum;
for(size_t i=minSeq; i<a.size(); ++i){
currentSum += a[i];
minimalSum += a[i] - a[i-minSeq];
if(currentSum < minimalSum) currentSum = minimalSum;
if(currentSum > maxSum) maxSum = currentSum;
}
return maxSum;
}
关于c++ - 指定最小长度的线性时间最大连续子序列求和算法,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/33293532/