实现合并排序算法的问题

Implementing merge sort algorithm issue

本文关键字：问题 算法 排序 合并 实现      更新时间：2023-10-16

我正在实现一个合并排序算法，在合并算法中收到一个std:：bad_alloc，使用cerr语句，我发现我的错误出现在合并算法的第一个循环中。然而，我不知道出了什么问题。

vector<int> VectorOps::mergeSort(vector<int> toSort)
{
if(toSort.size() <= 1)
{
return toSort;
}
vector<int> left;
vector<int> right;
int half = toSort.size()/2;
for(int i = 0; i < half; ++i)
{
left.push_back(toSort.at(i));
}
for(int i = half; i < toSort.size(); ++i)
{
right.push_back(toSort.at(i));
}
//merge algorithim
vector<int> toReturn;
while(left.size() > 0 || right.size() > 0)
{
cerr << "The numbers are "<< endl;
if(left.size() > 0 && right.size() > 0)
{
if(left.at(0) <= right.at(0))
{
toReturn.push_back(left.at(0));
}
else
{
toReturn.push_back(right.at(0));
}
}
else if(left.size() > 0)
{
toReturn.push_back(left.at(0));
}
else if(right.size() > 0)
{
toReturn.push_back(right.at(0));
}
}
return toReturn;
}

In:

while(left.size() > 0 || right.size() > 0)

left和right的大小永远不会改变(您不会移除head元素)，因此toReturn在没有绑定的情况下增长，并且您会耗尽内存。

正如@BenJackson在回答中已经提到的。。。

left和right的大小从未改变。你只是从向量中得到元素，而不是从中移除。因此，toReturn的大小在没有绑定的情况下增长。

vector没有任何去除头部元素的方法，但你可以像一样实现

left.erase(left.begin());

对于您的解决方案，要么从向量中移除head元素，要么只迭代向量并获取值。

vector<int> toReturn;
int l = 0; r = 0;
while (l < left.size() || r < right.size()) {
if (l < left.size() && r < right.size()) {
if (left.at(l) <= right.at(r)) {
toReturn.push_back(left.at(l++));
} else {
toReturn.push_back(right.at(r++));
}
} else if (l < left.size()) {
toReturn.push_back(left.at(l++));
} else if (r < right.size()) {
toReturn.push_back(right.at(r++));
}
}

通过擦除头元素合并实现。

while (left.size() > 0 || right.size() > 0) {
cerr << "The numbers are " << endl;
if (left.size() > 0 && right.size() > 0) {
if (left.at(0) <= right.at(0)) {
toReturn.push_back(left.at(0));
//erase head element from left
left.erase(left.begin());
} else {
toReturn.push_back(right.at(0));
//erase head element from right
right.erase(right.begin());
}
} else if (left.size() > 0) {
toReturn.push_back(left.at(0));
//erase head element from left
left.erase(left.begin());
} else if (right.size() > 0) {
toReturn.push_back(right.at(0));
//erase head element from right
right.erase(right.begin());
}
}
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