如果您对列表进行第一次遍历，并且仅提取可能有3项AP的数字对（差异为0 mod 2），则可以大大缩短执行时间。然后在这些对之间进行迭代。
伪C ++-y代码：
// Contains information about each beginning pointstruct BeginNode { int value; size_t offset; SortedList<EndNode> ends; //sorted by EndNode.value};// Contains information about each class of end pointstruct EndNode { int value; List<size_t> offsets; // will be sorted without effort due to how we collect offsets};struct Result { size_t begin; size_t middle; size_t end;};SortedList<BeginNode> nodeList;foreach (auto i : baseList) { BeginNode begin; node.value = i; node.offset = i's offset; //you'll need to use old school for (i=0;etc;i++) with this // baseList is the list between begin and end-2 (inclusive) foreach (auto j : restList) { // restList is the list between iterator i+2 and end (inclusive) // we do not need to consider i+1, because not enough space for AP if ((i-j)%2 == 0) { //if it's possible to have a 3 term AP between these two nodes size_t listOffset = binarySearch(begin.ends); if (listOffset is valid) { begin.ends[listOffset].offsets.push_back(offsets); } else { EndNode end; end.value = j; end.offsets.push_back(j's offset); begin.ends.sorted_insert(end); } } } if (begin has shit in it) { nodeList.sorted_insert(begin); }}// Collection done, now iterate over collectionList<Result> res;foreach (auto node : nodeList) { foreach (auto endNode : node.ends) { foreach (value : sublist from node.offset until endNode.offsets.last()) { if (value == average(node.value, endNode.value)) { // binary_search here to determine how many offsets in "endNode.offsets" "value's offset" is less than. do this that many times: res.push_back({node.value, value, endNode.value}); } } }}return res;
