Changeset 14970 in josm for trunk/src/org/openstreetmap
- Timestamp:
- 2019-04-07T19:03:21+02:00 (6 years ago)
- File:
-
- 1 edited
Legend:
- Unmodified
- Added
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trunk/src/org/openstreetmap/josm/actions/SimplifyWayAction.java
r14654 r14970 134 134 * @param way the way to be simplified 135 135 * @param node the node to check 136 * @param multipleUseNodes set of nodes which is used more than once in the way 136 137 * @return true if <code>node</code> is a required node which can't be removed 137 138 * in order to simplify the way. 138 139 */ 139 protected static boolean isRequiredNode(Way way, Node node) { 140 protected static boolean isRequiredNode(Way way, Node node, Set<Node> multipleUseNodes) { 140 141 boolean isRequired = node.isTagged(); 141 if (!isRequired) { 142 if (!isRequired && multipleUseNodes.contains(node)) { 142 143 int frequency = Collections.frequency(way.getNodes(), node); 143 144 if ((way.getNode(0) == node) && (way.getNode(way.getNodesCount()-1) == node)) { … … 164 165 public final SequenceCommand simplifyWay(Way w) { 165 166 return simplifyWay(w, Config.getPref().getDouble("simplify-way.max-error", 3.0)); 167 } 168 169 /** 170 * Calculate a set of nodes which occurs more than once in the way 171 * @param w the way 172 * @return a set of nodes which occurs more than once in the way 173 */ 174 private static Set<Node> getMultiUseNodes(Way w) { 175 Set<Node> multipleUseNodes = new HashSet<>(); 176 Set<Node> allNodes = new HashSet<>(); 177 for (Node n : w.getNodes()) { 178 if (!allNodes.add(n)) 179 multipleUseNodes.add(n); 180 } 181 return multipleUseNodes; 166 182 } 167 183 … … 177 193 int lower = 0; 178 194 int i = 0; 195 196 Set<Node> multipleUseNodes = getMultiUseNodes(w); 179 197 List<Node> newNodes = new ArrayList<>(w.getNodesCount()); 180 198 while (i < w.getNodesCount()) { 181 if (isRequiredNode(w, w.getNode(i))) { 199 if (isRequiredNode(w, w.getNode(i), multipleUseNodes)) { 182 200 // copy a required node to the list of new nodes. Simplify not possible 183 201 newNodes.add(w.getNode(i)); … … 188 206 i++; 189 207 // find the longest sequence of not required nodes ... 190 while (i < w.getNodesCount() && !isRequiredNode(w, w.getNode(i))) { 208 while (i < w.getNodesCount() && !isRequiredNode(w, w.getNode(i), multipleUseNodes)) { 191 209 i++; 192 210 } … … 198 216 199 217 // Closed way, check if the first node could also be simplified ... 200 if (newNodes.size() > 3 && newNodes.get(0) == newNodes.get(newNodes.size() - 1) && !isRequiredNode(w, newNodes.get(0))) { 218 if (newNodes.size() > 3 && newNodes.get(0) == newNodes.get(newNodes.size() - 1) && !isRequiredNode(w, newNodes.get(0), multipleUseNodes)) { 201 219 final List<Node> l1 = Arrays.asList(newNodes.get(newNodes.size() - 2), newNodes.get(0), newNodes.get(1)); 202 220 final List<Node> l2 = new ArrayList<>(3);
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