LeetCode刷题总结--DFS与字符串处理
388. Longest Absolute File Path
"dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"这个字符串序列是由文件目录树的深度优先搜索（先根遍历）得到的，所以需要借助栈来进行解析这个字符串得到所有全路径，从而比较全路径字符串的长度。这里要注意的是一个子文件夹下可能还包含子文件夹和文件，所以借助单调栈（每个当前相对路径的深度或层次是单调递增的）。
import javafx.util.Pair;
import java.util.LinkedList;
import java.util.Stack;
public class LongestAbsoluteFilePath {
public static int lengthLongestPath(String input) {
String[] lists = input.split("\n");
LinkedList> table = new LinkedList<>();
for (String s : lists) {
int count = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '\t')
count++;
else
break;
}
Pair tmp = new Pair<>(s.substring(count), count);
table.add(tmp);
}
Stack> stack = new Stack<>();
int ret = 0;
int curr = 0;
int wc = 0;
for (Pair pair : table) {
String key = (String) pair.getKey();
Integer val = (Integer) pair.getValue();
while (!stack.isEmpty() && stack.peek().getValue() >= val) {
String s = stack.pop().getKey();
curr -= s.length();
wc -= 1;
}
stack.add(pair);
curr += key.length();
wc += 1;
if (key.contains("."))
ret = Math.max(ret, curr + wc - 1);
}
return ret;
}
public static void main(String[] args) {
String ex="dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext";
System.out.println(lengthLongestPath(ex));
String[] tmp=ex.split("\n");
for (String s:tmp)
System.out.println(s);
}
}
386. Lexicographical Numbers
算法一：字符串排序O(n+nlogn)
class Solution {
public List lexicalOrder(int n) {
LinkedList list=new LinkedList<>();
String[] vals=new String[n];
for(int i=1;i<=n;i++)
vals[i-1]=String.valueOf(i);
Arrays.sort(vals);
for(String val:vals)
list.add(Integer.valueOf(val));
return list;
}
}
算法二：栈
class Solution {
public List lexicalOrder(int n) {
LinkedList ret = new LinkedList<>();
Stack stack = new Stack<>();
for (int i = n <= 9 ? n : 9; i > 0; i--) {
stack.push(i);
}
while (!stack.isEmpty()){
int cur=stack.pop();
ret.add(cur);
for(int j=9;j>=0;j--){
int val=cur*10+j;
if(val<=n)
stack.push(val);
}
}
return ret;
}
}
算法三：DFS,这种解法特别优美
class Solution {
public static LinkedList ret;
public List lexicalOrder(int n) {
ret=new LinkedList<>();
for(int i=1;i<=9;i++)
{
if(i<=n)
{
ret.add(i);
dfs(i,n);
}
}
return ret;
}
public static void dfs(int num,int n){
if(num>n)
return;
for(int i=0;i<=9;i++)
{
int t=num*10+i;
if(t<=n){
ret.add(t);
dfs(t,n);
}
else
break;
}
}
}
208. Implement Trie (Prefix Tree)
在算法导论上看到这种数据结构，真的优美极了！
class Trie {
private TrieNode root;
class TrieNode{
private TrieNode[] links;
private final int R=26;
private boolean isEnd;
public TrieNode(){
links=new TrieNode[R];
}
public boolean containsKey(char c){
return links[c-'a']!=null;
}
public TrieNode get(char c){
return links[c-'a'];
}
public void put(char c,TrieNode node){
links[c-'a']=node;
}
public void setEnd(){
isEnd=true;
}
}
/** Initialize your data structure here. */
public Trie() {
root=new TrieNode();
}
/** Inserts a word into the trie. */
public void insert(String word) {
TrieNode node=root;
for(int i=0;i211. Add and Search Word - Data structure design
class WordDictionary {
public TrieNode root;
class TrieNode {
private TrieNode[] links;
private final int R = 26;
private boolean isEnd;
public TrieNode() {
links = new TrieNode[R];
}
public boolean containsKey(char c) {
return links[c - 'a'] != null;
}
public TrieNode get(char c) {
return links[c - 'a'];
}
public void put(char c, TrieNode node) {
links[c - 'a'] = node;
}
public void setEnd() {
isEnd = true;
}
}
/** Initialize your data structure here. */
public WordDictionary() {
root = new TrieNode();
}
/** Adds a word into the data structure. */
public void addWord(String word) {
TrieNode node = this.root;
for (int i = 0; i < word.length(); i++) {
char curChar = word.charAt(i);
if (!node.containsKey(curChar))
node.put(curChar, new TrieNode());
node = node.get(curChar);
}
node.isEnd = true;
}
private boolean search(TrieNode node, String word, int idx) {
if (idx == word.length())
return node.isEnd;
char c = word.charAt(idx);
boolean flag = false;
if (c == '.') {
for (TrieNode tmp : node.links) {
if (tmp != null)
flag = flag || search(tmp, word, idx + 1);
}
} else if (node.get(c) == null){
return false;
}
else
return search(node.get(c), word, idx + 1);
return flag;
}
/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
public boolean search(String word) {
return search(this.root, word, 0);
}
}
/**
* Your WordDictionary object will be instantiated and called as such:
* WordDictionary obj = new WordDictionary();
* obj.addWord(word);
* boolean param_2 = obj.search(word);
*/