C++ 合并排序算法的实现
我正在实现一个合并排序算法,并且在合并算法中收到一个std::bad_alloc,使用cerr语句,我发现我的错误在合并算法的第一个循环中。然而,我无法找出问题所在C++ 合并排序算法的实现,c++,C++,我正在实现一个合并排序算法,并且在合并算法中收到一个std::bad_alloc,使用cerr语句,我发现我的错误在合并算法的第一个循环中。然而,我无法找出问题所在 vector<int> VectorOps::mergeSort(vector<int> toSort) { if(toSort.size() <= 1) { return toSort; } vector<int> left; ve
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;
}
向量向量操作::合并排序(向量到排序)
{
如果(toSort.size()0 | | right.size()>0)
{
cerr(0)
{
if(左,在(0)0处)
{
toReturn.向后推_(在(0)处向左);
}
else if(right.size()>0)
{
返回。向后推(右。在(0)处);
}
}
回归回归;
}
在:
left
和right
的大小永远不会改变(您不会删除head元素),因此toreern
的大小会在没有绑定的情况下增长,并且内存会耗尽。正如@BenJackson在回答中已经提到的那样
left
和right
的大小永远不会改变。您只是从向量中获取元素,而不是从向量中移除。因此,toReturn
的大小不受约束地增长
vector
没有任何方法来删除头部元素,但您可以实现类似的功能
left.erase(left.begin());
对于您的解决方案,要么从vector中删除head元素,要么只迭代vector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++));
}
}
vectoreturn;
int l=0;r=0;
而(l0){
cerr(0){
if(左,在(0)0处){
toReturn.向后推_(在(0)处向左);
//从左侧删除头部元素
left.erase(left.begin());
}else if(right.size()>0){
返回。向后推(右。在(0)处);
//从右侧删除头部元素
right.erase(right.begin());
}
}
我建议您开始使用调试器,因为cerr
不允许您单步执行并查看变量。
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++));
}
}
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());
}
}