你的问题是对排名的误解。数组索引为size_t
not float
,所以你需要返回一个vector<size_t>
not a vector<float>
.
That said your sort is O(n2). If you're willing to use more memory we can get that time down to O(n log(n)):
vector<size_t> rankSort(const float* v_temp, const size_t size) {
vector<pair<float, size_t> > v_sort(size);
for (size_t i = 0U; i < size; ++i) {
v_sort[i] = make_pair(v_temp[i], i);
}
sort(v_sort.begin(), v_sort.end());
pair<double, size_t> rank;
vector<size_t> result(size);
for (size_t i = 0U; i < size; ++i) {
if (v_sort[i].first != rank.first) {
rank = make_pair(v_sort[i].first, i);
}
result[v_sort[i].second] = rank.second;
}
return result;
}
Live Example
EDIT:
是的,这实际上变得更简单一点vector<float>
代替float[]
:
vector<size_t> rankSort(const vector<float>& v_temp) {
vector<pair<float, size_t> > v_sort(v_temp.size());
for (size_t i = 0U; i < v_sort.size(); ++i) {
v_sort[i] = make_pair(v_temp[i], i);
}
sort(v_sort.begin(), v_sort.end());
pair<double, size_t> rank;
vector<size_t> result(v_temp.size());
for (size_t i = 0U; i < v_sort.size(); ++i) {
if (v_sort[i].first != rank.first) {
rank = make_pair(v_sort[i].first, i);
}
result[v_sort[i].second] = rank.second;
}
return result;
}
Live Example