[英]Finding the longest absolute file path using java
我實現了以下方法來查找最長的絕對文件路徑。
public static int lengthLongestPath(String input) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
if (input.length() == 0) return 0;
int maxLength = 0;
int subStringLength = 0;
int previousLevel = 0;
String[] paths = input.split("\n");
for (String path : paths) {
String[] substr = path.split("\t");
String dirOrFile = substr[substr.length-1];
int level = substr.length -1;
if(level <= previousLevel && level !=0){
previousLevel = level-1;
subStringLength = map.get(previousLevel);
}else if(level ==0){
subStringLength = 0;
}else{
previousLevel = level;
}
subStringLength += dirOrFile.length();
if(dirOrFile.contains(".")){
maxLength = Math.max(subStringLength+level, maxLength);
}
map.put(level, subStringLength);
}
return maxLength;
}
但是,以下輸入失敗了。 預期的輸出是528.但我的輸出是549。
"sladjf\n\tlkjlkv\n\t\tlkjlakjlert\n\t\t\tlaskjglaksjf\n\t\t\t\tlakjgfljrtlj\n\t\t\t\t\tlskajflakjsvlj\n\t\t\t\t\t\tlskgjflkjrtlrjt\n\t\t\t\t\t\t\tlkjglkjlvkjdlvkj\n\t\t\t\t\t\t\t\tlfjkglkjfljdlv\n\t\t\t\t\t\t\t\t\tlkdfjglerjtkrjkljsd.lkvjlkajlfk\n\t\t\t\t\t\t\tlskfjlksjljslvjxjlvkzjljajoiwjejlskjslfj.slkjflskjldfkjoietruioskljfkljf\n\t\t\t\t\tlkasjfljsaljlxkcjzljvl.asljlksaj\n\t\t\t\tasldjflksajf\n\t\t\t\talskjflkasjlvkja\n\t\t\t\twioeuoiwutrljsgfjlskfg\n\t\t\t\taslkjvlksjvlkjsflgj\n\t\t\t\t\tlkvnlksfgk.salfkjaslfjskljfv\n\t\t\tlksdjflsajlkfj\n\t\t\tlasjflaskjlk\n\t\tlsakjflkasjfkljas\n\t\tlskjvljvlkjlsjfkgljfg\n\tsaljkglksajvlkjvkljlkjvksdj\n\tlsakjglksajkvjlkjdklvj\n\tlskjflksjglkdjbkljdbkjslkj\n\t\tlkjglkfjkljsdflj\n\t\t\tlskjfglkjdfgkljsdflj\n\t\t\t\tlkfjglksdjlkjbsdlkjbk\n\t\t\t\t\tlkfgjlejrtljkljsdflgjl\n\t\t\t\t\tsalgkfjlksfjgkljsgfjl\n\t\t\t\t\tsalkflajwoieu\n\t\t\t\t\t\tlaskjfglsjfgljkkvjsdlkjbklds\n\t\t\t\t\t\t\tlasjglriotuojgkjsldfgjsklfgjl\n\t\t\t\t\t\t\t\tlkajglkjskljsdljblkdfjblfjlbjs\n\t\t\t\t\t\t\t\t\tlkajgljroituksfglkjslkjgoi\n\t\t\t\t\t\t\t\t\t\tlkjglkjkljkljdkbljsdfljgklfdj\n\t\t\t\t\t\t\t\t\t\t\tlkjlgkjljgslkdkldjblkj\n\t\t\t\t\t\t\t\t\t\t\t\tlkjfglkjlkjbsdklj.slgfjalksjglkfjglf\n\t\t\t\t\t\t\t\t\t\t\t\tlkasjrlkjwlrjljsl\n\t\t\t\t\t\t\t\t\t\t\t\t\tlksjgflkjfklgjljbljls\n\t\t\t\t\t\t\t\t\t\t\t\t\t\tlkjsglkjlkjfkljdklbjkldf\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\tlkjglkjdlsfjdglsdjgjlxljjlrjsgjsjlk\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\tlkjsgkllksjfgjljdslfkjlkasjdflkjxcljvlkjsgkljsfg\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\tlaskjlkjsakljglsdjfgksdjlkgjdlskjb\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\tlkajsgfljfklgjlkdjgfklsdjklj\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\tlkjfglkjlkgjlkjl.aslkjflasjlajglkjaf\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\tlkjasflgjlskjglkfjgklgsdjflkbjsdklfjskldfjgklsfdjgklfdjgl\n\tlskadjlkjsldwwwwwfj\n\t\tlkjflkasjlfjlkjajslfkjlasjkdlfjlaskjalvwwwwwwwwwwwwwwwkjlsjfglkjalsjgflkjaljlkdsjslbjsljksldjlsjdlkjljvblkjlkajfljgasljfkajgfljfjgldjblkjsdljgsldjg.skljf"
有人可以幫助我找到我的代碼的問題。
給出關於字符串格式的一些背景以下字符串表示:以下文件路徑。
"dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"
dir
subdir1
file1.ext
subsubdir1
subdir2
subsubdir2
file2.ext
簡化現有代碼將提供以下內容,它們會為您的兩個示例返回正確的結果:
public static int lengthLongestPath(String input) {
if (input.length() == 0)
return 0;
Map<Integer, Integer> map = new HashMap<>();
int maxLength = 0;
String[] paths = input.split("\n");
for (String path : paths) {
String dirOrFile = path.replaceFirst("\\t*", "");
int level = path.lastIndexOf("\t") + 1;
int prefixLength = 0;
if (level > 0) {
prefixLength = map.get(level - 1);
}
int pathLength = prefixLength + dirOrFile.length();
if (dirOrFile.contains(".")) {
maxLength = Math.max(pathLength + level, maxLength);
}
map.put(level, pathLength);
}
return maxLength;
}
我想你可能會讓事情變得過於復雜。 由於字符串的格式,您無法“返回”到您已經離開的目錄,因此您無需保留地圖。 另一種實現可能是用\\n
字符拆分字符串並使用列表來跟蹤您在路徑中的位置,其中\\t
字符數表示您需要保留的位置:
public static int lengthLongestPath(String input) {
int maxLen = 0;
List<String> currPath = new LinkedList<>();
String[] parts = input.split("\n");
for (String part : parts) {
int numTabs = Math.max(0, part.lastIndexOf('\t'));
part = part.substring(numTabs + 1);
currPath = currPath.subList(0, numTabs);
currPath.add(part);
int curLen = currPath.stream().mapToInt(String::length).sum();
maxLen = Math.max(maxLen, curLen);
}
return maxLen;
}
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