前缀树
前缀树 - LeetBook - 力扣（LeetCode）全球极客挚爱的技术成长平台
实现Trie（前缀树）
思路
书本也有官解，使用的HashMap，下面这个思路更快，使用数组
实现 Trie (前缀树) - 实现 Trie (前缀树) - 力扣（LeetCode）
解法1
class Trie {
private Trie[] children;
private boolean isEnd;
public Trie() {
children = new Trie[26];
isEnd = false;
}
public void insert(String word) {
Trie node = this;
for (int i = 0; i < word.length(); i++) {
char ch = word.charAt(i);
int index = ch - 'a';
if (node.children[index] == null) {
node.children[index] = new Trie();
}
node = node.children[index];
}
node.isEnd = true;
}
public boolean search(String word) {
Trie node = searchPrefix(word);
return node != null && node.isEnd;
}
public boolean startsWith(String prefix) {
return searchPrefix(prefix) != null;
}
public Trie searchPrefix(String prefix) {
Trie node = this;
for (int i = 0; i < prefix.length(); i++) {
char ch = prefix.charAt(i);
int index = ch - 'a';
if (node.children[index] == null) {
return null;
}
node = node.children[index];
}
return node;
}
}
/**
* Your Trie object will be instantiated and called as such:
* Trie obj = new Trie();
* obj.insert(word);
* boolean param_2 = obj.search(word);
* boolean param_3 = obj.startsWith(prefix);
*/
键值映射
官解思路写挺全
解法1
直接暴力丢到HashMap中
class MapSum {
Map<String, Integer> map;
public MapSum() {
map = new HashMap<>();
}
public void insert(String key, int val) {
map.put(key, val);
}
public int sum(String prefix) {
int result = 0;
for (String s : map.keySet()) {
if (s.startsWith(prefix)) {
result += map.get(s);
}
}
return result;
}
}
/**
* Your MapSum object will be instantiated and called as such:
* MapSum obj = new MapSum();
* obj.insert(key,val);
* int param_2 = obj.sum(prefix);
*/
单词替换
解法1
复用一下实现Trie这道题的Trie的结构
class Solution {
public String replaceWords(List<String> dictionary, String sentence) {
Trie trie = new Trie();
for (String dict : dictionary) {
trie.insert(dict);
}
StringBuilder result = new StringBuilder();
for (String sen : sentence.split(" ")) {
if (result.length() > 0) {
result.append(" ");
}
result.append(trie.search(sen));
}
return result.toString();
}
}
class Trie {
private Trie[] children;
private String pre;
public Trie() {
children = new Trie[26];
}
public void insert(String word) {
Trie node = this;
for (int i = 0; i < word.length(); i++) {
char ch = word.charAt(i);
int index = ch - 'a';
if (node.children[index] == null) {
node.children[index] = new Trie();
}
node = node.children[index];
}
node.pre = word;
}
public String search(String prefix) {
Trie node = this;
for (int i = 0; i < prefix.length(); i++) {
char ch = prefix.charAt(i);
int index = ch - 'a';
if (node == null) {
break;
}
if (node.pre != null) {
return node.pre;
}
node = node.children[index];
}
return prefix;
}
}
添加与搜索单词-数据结构设计
解法1
class WordDictionary {
private Trie root;
public WordDictionary() {
root = new Trie();
}
public void addWord(String word) {
root.insert(word);
}
public boolean search(String word) {
return dfs(word, 0, root);
}
public boolean dfs(String word, int index, Trie node) {
if (index == word.length()) {
return node.isEnd;
}
char ch = word.charAt(index);
if (Character.isLetter(ch)) {
int childIndex = ch - 'a';
Trie child = node.getChildren()[childIndex];
if (child != null && dfs(word, index + 1, child)) {
return true;
}
} else {
for (int i = 0; i < 26; i++) {
Trie child = node.getChildren()[i];
if (child != null && dfs(word, index + 1, child)) {
return true;
}
}
}
return false;
}
}
class Trie {
private Trie[] children;
public boolean isEnd;
public Trie() {
children = new Trie[26];
isEnd = false;
}
public void insert(String word) {
Trie node = this;
for (int i = 0; i < word.length(); i++) {
char ch = word.charAt(i);
int index = ch - 'a';
if (node.children[index] == null) {
node.children[index] = new Trie();
}
node = node.children[index];
}
node.isEnd = true;
}
public Trie[] getChildren() {
return children;
}
public boolean isEnd() {
return isEnd;
}
}
/**
* Your WordDictionary object will be instantiated and called as such:
* WordDictionary obj = new WordDictionary();
* obj.addWord(word);
* boolean param_2 = obj.search(word);
*/
数组中两个数的最大异或值
尝试暴力，被劝退😂
思路
宫水三叶大佬思路
【宫水三叶の相信科学系列】详解为何能用「贪心」&「Trie」找「最大异或结果」 - 数组中两个数的最大异或值 - 力扣（LeetCode）
解法1
class Solution {
class Node {
Node[] ns = new Node[2];
}
Node root = new Node();
void add(int x) {
Node p = root;
for (int i = 31; i >= 0; i--) {
int u = (x >> i) & 1;
if (p.ns[u] == null) {
p.ns[u] = new Node();
}
p = p.ns[u];
}
}
int getVal(int x) {
int ans = 0;
Node p = root;
for (int i = 31; i >= 0; i--) {
int a = (x >> i) & 1;
int b = 1 - a;
if (p.ns[b] != null) {
ans |= (b << i);
p = p.ns[b];
} else {
ans |= (a << i);
p = p.ns[a];
}
}
return ans;
}
public int findMaximumXOR(int[] nums) {
int ans = 0;
for (int i : nums) {
add(i);
int j = getVal(i);
ans = Math.max(ans, i ^ j);
}
return ans;
}
}
单词搜索Ⅱ
思路
三叶大佬思路
【宫水三叶】一题双解 :「回溯」&「Trie 」 - 单词搜索 II - 力扣（LeetCode）
解法1
class Solution {
class TrieNode {
String s;
TrieNode[] tns = new TrieNode[26];
}
void insert(String s) {
TrieNode p = root;
for (int i = 0; i < s.length(); i++) {
int u = s.charAt(i) - 'a';
if (p.tns[u] == null) p.tns[u] = new TrieNode();
p = p.tns[u];
}
p.s = s;
}
Set<String> set = new HashSet<>();
char[][] board;
int n, m;
TrieNode root = new TrieNode();
int[][] dirs = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
boolean[][] vis = new boolean[15][15];
public List<String> findWords(char[][] _board, String[] words) {
board = _board;
m = board.length; n = board[0].length;
for (String w : words) insert(w);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int u = board[i][j] - 'a';
if (root.tns[u] != null) {
vis[i][j] = true;
dfs(i, j, root.tns[u]);
vis[i][j] = false;
}
}
}
List<String> ans = new ArrayList<>();
for (String s : set) ans.add(s);
return ans;
}
void dfs(int i, int j, TrieNode node) {
if (node.s != null) set.add(node.s);
for (int[] d : dirs) {
int dx = i + d[0], dy = j + d[1];
if (dx < 0 || dx >= m || dy < 0 || dy >= n) continue;
if (vis[dx][dy]) continue;
int u = board[dx][dy] - 'a';
if (node.tns[u] != null) {
vis[dx][dy] = true;
dfs(dx, dy, node.tns[u]);
vis[dx][dy] = false;
}
}
}
}
回文对
思路
官解
温馨提示：前缀树会超时
解法1
class Solution {
List<String> wordsRev = new ArrayList<String>();
Map<String, Integer> indices = new HashMap<String, Integer>();
public List<List<Integer>> palindromePairs(String[] words) {
int n = words.length;
for (String word: words) {
wordsRev.add(new StringBuffer(word).reverse().toString());
}
for (int i = 0; i < n; ++i) {
indices.put(wordsRev.get(i), i);
}
List<List<Integer>> ret = new ArrayList<List<Integer>>();
for (int i = 0; i < n; i++) {
String word = words[i];
int m = words[i].length();
if (m == 0) {
continue;
}
for (int j = 0; j <= m; j++) {
if (isPalindrome(word, j, m - 1)) {
int leftId = findWord(word, 0, j - 1);
if (leftId != -1 && leftId != i) {
ret.add(Arrays.asList(i, leftId));
}
}
if (j != 0 && isPalindrome(word, 0, j - 1)) {
int rightId = findWord(word, j, m - 1);
if (rightId != -1 && rightId != i) {
ret.add(Arrays.asList(rightId, i));
}
}
}
}
return ret;
}
public boolean isPalindrome(String s, int left, int right) {
int len = right - left + 1;
for (int i = 0; i < len / 2; i++) {
if (s.charAt(left + i) != s.charAt(right - i)) {
return false;
}
}
return true;
}
public int findWord(String s, int left, int right) {
return indices.getOrDefault(s.substring(left, right + 1), -1);
}
}
