我正在hackerRank上解决此question。我遍历了整个讨论部分,尝试了所有带有预期结果的建议测试用例。我想我可能在做一些愚蠢的代码错误,因为我确定我已经考虑/考虑了实现中的每种情况。如果我的代码有任何错误,请您帮我指出一下。
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String strNum[] = bf.readLine().split("\\s");
double n = Double.parseDouble(strNum[0]);
double m = Double.parseDouble(strNum[1]);
double k = Double.parseDouble(strNum[2]);
Map<Double, TreeMap<Double, Double>> map = new HashMap<>();
while (k > 0) {
strNum = bf.readLine().split("\\s");
double r = Double.parseDouble(strNum[0]);
double c1 = Double.parseDouble(strNum[1]);
double c2 = Double.parseDouble(strNum[2]);
TreeMap<Double, Double> innerMap = map.get(r);
if (innerMap != null) {
Double x = innerMap.get(c1);
if (x != null) {
if (c2 > x) {
innerMap.put(c1, c2);
}
} else {
innerMap.put(c1, c2);
}
} else {
innerMap = new TreeMap<Double, Double>();
innerMap.put(c1, c2);
map.put(r, innerMap);
}
k--;
}
double count = (n - map.size()) * m;
for (Map.Entry<Double, TreeMap<Double, Double>> e : map.entrySet()) {
TreeMap<Double, Double> innerMap = e.getValue();
double start = innerMap.firstKey();
double end = innerMap.firstEntry().getValue();
for (Map.Entry<Double, Double> e2 : innerMap.entrySet()) {
double x = e2.getKey();
double y = e2.getValue();
if (y > end) {
if (x > end) {
count += ((x - end) - 1);
}
end = y;
}
}
count += (m - (end - start + 1));
}
System.out.println(String.format("%.0f", count));
}
24/31测试用例失败。非常感谢您的帮助。
如果您的代码读取的是r,c1,c2
,然后是r,c1,c2'
和c2'<c2
,则它将静默地删除前一首曲目。