我在地图上绘制struct const_node
{
const_node(char content, const_node* next)
: next(next), content(content)
{
}
const_node* const next;
const char content;
};
// put front in front of each existing sub-result
void update_helper(int front, std::vector<const_node*>& intermediate)
{
for (size_t i = 0; i < intermediate.size(); i++)
{
intermediate[i] = new const_node(front + 'a', intermediate[i]);
}
if (intermediate.empty())
{
intermediate.push_back(new const_node(front + 'a', NULL));
}
}
std::vector<const_node*> decode_it(int digits[9], size_t count)
{
int current = 0;
std::vector<const_node*> intermediates[3];
for (size_t i = 0; i < count; i++)
{
current = (current + 1) % 3;
int prev = (current + 2) % 3; // -1
int prevprev = (current + 1) % 3; // -2
size_t index = count - i - 1; // invert direction
// copy from prev
intermediates[current] = intermediates[prev];
// update current (part 1)
update_helper(digits[index], intermediates[current]);
if (index + 1 < count && digits[index] &&
digits[index] * 10 + digits[index + 1] < 26)
{
// update prevprev
update_helper(digits[index] * 10 + digits[index + 1], intermediates[prevprev]);
// add to current (part 2)
intermediates[current].insert(intermediates[current].end(), intermediates[prevprev].begin(), intermediates[prevprev].end());
}
}
return intermediates[current];
}
void cleanupDelete(std::vector<const_node*>& nodes);
int main()
{
int code[] = { 1, 2, 3, 1, 2, 3, 1, 2, 3 };
int size = sizeof(code) / sizeof(int);
std::vector<const_node*> result = decode_it(code, size);
// output
for (size_t i = 0; i < result.size(); i++)
{
std::cout.width(3);
std::cout.flags(std::ios::right);
std::cout << i << ": ";
const_node* item = result[i];
while (item)
{
std::cout << item->content;
item = item->next;
}
std::cout << std::endl;
}
cleanupDelete(result);
}
void fillCleanup(const_node* n, std::set<const_node*>& all_nodes)
{
if (n)
{
all_nodes.insert(n);
fillCleanup(n->next, all_nodes);
}
}
void cleanupDelete(std::vector<const_node*>& nodes)
{
// this is like multiple inverse trees, hard to delete correctly, since multiple next pointers refer to the same target
std::set<const_node*> all_nodes;
for each (auto var in nodes)
{
fillCleanup(var, all_nodes);
}
nodes.clear();
for each (auto var in all_nodes)
{
delete var;
}
all_nodes.clear();
}
就像动画一样。如下。
polyline如何删除m_handler = new Handler();
m_handlerTask = new Runnable() {
@Override
public void run() {
//line.remove();
if (t < pointsPoly.size() - 1) {
LatLng src = pointsPoly.get(t);
LatLng dest = pointsPoly.get(t + 1);
Polyline lineAnimation = googleMap.addPolyline(new PolylineOptions()
.add(new LatLng(src.latitude, src.longitude),
new LatLng(dest.latitude, dest.longitude))
.width(10).color(Color.DKGRAY).geodesic(true));
t++;
} else {
t = 0;
}
m_handler.postDelayed(m_handlerTask, polyLineTimer);
}
};
m_handler.post(m_handlerTask);
？我不想polyline。
我试过了clearMap()，但它没有用。
答案 0 :(得分：4)
您只需执行以下操作，但不是将结果分配给变量，而是将其放入ArrayList ...
ArrayList<Polyline> lines = new ArrayList<>();
//Add line to map
lines.add(mMap.addPolyline(new PolylineOptions()
.add(new LatLng(location.getLatitude(), location.getLongitude()),
new LatLng(this.destinationLatitude, this.destinationLongitude))
.width(1)
.color(Color.DKGRAY));
//Remove the same line from map
line.remove();
从地图中删除此折线。删除折线后， 所有方法的行为都是未定义的。