feat: update solutions to lc/lcci problems

* lc No.0139.Word Break
* lcci No.17.15.Longest Word

Author: yanglbme

lcci/17.15.Longest Word/README.md

@@ -62,12 +62,7 @@ class Solution:
             return -1 if a > b else 1
 
         def dfs(w):
-            if not w:
-                return True
-            for i in range(1, len(w) + 1):
-                if trie.search(w[:i]) and dfs(w[i:]):
-                    return True
-            return False
+            return not w or any(trie.search(w[:i]) and dfs(w[i:]) for i in range(1, len(w) + 1))
 
         words.sort(key=cmp_to_key(cmp))
         trie = Trie()

lcci/17.15.Longest Word/README_EN.md

@@ -64,12 +64,7 @@ class Solution:
             return -1 if a > b else 1
 
         def dfs(w):
-            if not w:
-                return True
-            for i in range(1, len(w) + 1):
-                if trie.search(w[:i]) and dfs(w[i:]):
-                    return True
-            return False
+            return not w or any(trie.search(w[:i]) and dfs(w[i:]) for i in range(1, len(w) + 1))
 
         words.sort(key=cmp_to_key(cmp))
         trie = Trie()

lcci/17.15.Longest Word/Solution.py

@@ -30,12 +30,7 @@ def cmp(a, b):
             return -1 if a > b else 1
 
         def dfs(w):
-            if not w:
-                return True
-            for i in range(1, len(w) + 1):
-                if trie.search(w[:i]) and dfs(w[i:]):
-                    return True
-            return False
+            return not w or any(trie.search(w[:i]) and dfs(w[i:]) for i in range(1, len(w) + 1))
 
         words.sort(key=cmp_to_key(cmp))
         trie = Trie()

solution/0100-0199/0139.Word Break/README.md

@@ -52,9 +52,11 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-动态规划法。
+**方法一：动态规划**
 
-`dp[i]` 表示前 i 个字符组成的字符串 `s[0...i-1]` 能否拆分成若干个字典中出现的单词。
+$dp[i]$ 表示前 $i$ 个字符组成的字符串 $s[0...i-1]$ 能否拆分成若干个字典中出现的单词。
+
+时间复杂度 $O(n^2)$。
 
 <!-- tabs:start -->
 
@@ -74,7 +76,7 @@ class Solution:
                 if dp[j] and s[j:i] in words:
                     dp[i] = True
                     break
-        return dp[n]
+        return dp[-1]
 ```
 
 ### **Java**
@@ -107,16 +109,16 @@ class Solution {
 class Solution {
 public:
     bool wordBreak(string s, vector<string>& wordDict) {
-        unordered_set<string> words;
-        for (auto word : wordDict) {
-            words.insert(word);
-        }
+        unordered_set<string> words(wordDict.begin(), wordDict.end());
         int n = s.size();
-        vector<bool> dp(n + 1, false);
+        vector<bool> dp(n + 1);
         dp[0] = true;
-        for (int i = 1; i <= n; ++i) {
-            for (int j = 0; j < i; ++j) {
-                if (dp[j] && words.find(s.substr(j, i - j)) != words.end()) {
+        for (int i = 1; i <= n; ++i)
+        {
+            for (int j = 0; j < i; ++j)
+            {
+                if (dp[j] && words.count(s.substr(j, i - j)))
+                {
                     dp[i] = true;
                     break;
                 }

solution/0100-0199/0139.Word Break/README_EN.md

@@ -64,7 +64,7 @@ class Solution:
                 if dp[j] and s[j:i] in words:
                     dp[i] = True
                     break
-        return dp[n]
+        return dp[-1]
 ```
 
 ### **Java**
@@ -95,16 +95,16 @@ class Solution {
 class Solution {
 public:
     bool wordBreak(string s, vector<string>& wordDict) {
-        unordered_set<string> words;
-        for (auto word : wordDict) {
-            words.insert(word);
-        }
+        unordered_set<string> words(wordDict.begin(), wordDict.end());
         int n = s.size();
-        vector<bool> dp(n + 1, false);
+        vector<bool> dp(n + 1);
         dp[0] = true;
-        for (int i = 1; i <= n; ++i) {
-            for (int j = 0; j < i; ++j) {
-                if (dp[j] && words.find(s.substr(j, i - j)) != words.end()) {
+        for (int i = 1; i <= n; ++i)
+        {
+            for (int j = 0; j < i; ++j)
+            {
+                if (dp[j] && words.count(s.substr(j, i - j)))
+                {
                     dp[i] = true;
                     break;
                 }

solution/0100-0199/0139.Word Break/Solution.cpp

@@ -1,16 +1,16 @@
 class Solution {
 public:
     bool wordBreak(string s, vector<string>& wordDict) {
-        unordered_set<string> words;
-        for (auto word : wordDict) {
-            words.insert(word);
-        }
+        unordered_set<string> words(wordDict.begin(), wordDict.end());
         int n = s.size();
-        vector<bool> dp(n + 1, false);
+        vector<bool> dp(n + 1);
         dp[0] = true;
-        for (int i = 1; i <= n; ++i) {
-            for (int j = 0; j < i; ++j) {
-                if (dp[j] && words.find(s.substr(j, i - j)) != words.end()) {
+        for (int i = 1; i <= n; ++i)
+        {
+            for (int j = 0; j < i; ++j)
+            {
+                if (dp[j] && words.count(s.substr(j, i - j)))
+                {
                     dp[i] = true;
                     break;
                 }

solution/0100-0199/0139.Word Break/Solution.cs

@@ -4,9 +4,12 @@ public bool WordBreak(string s, IList<string> wordDict) {
         int n = s.Length;
         var dp = new bool[n + 1];
         dp[0] = true;
-        for (int i = 1; i <= n; ++i) {
-            for (int j = 0; j < i; ++j) {
-                if (dp[j] && words.Contains(s.Substring(j, i))) {
+        for (int i = 1; i <= n; ++i)
+        {
+            for (int j = 0; j < i; ++j)
+            {
+                if (dp[j] && words.Contains(s.Substring(j, i - j)))
+                {
                     dp[i] = true;
                     break;
                 }

solution/0100-0199/0139.Word Break/Solution.java

@@ -1,15 +1,12 @@
-public class Solution {
-    public bool WordBreak(string s, IList<string> wordDict) {
-        var words = new HashSet<string>(wordDict);
-        int n = s.Length;
-        var dp = new bool[n + 1];
+class Solution {
+    public boolean wordBreak(String s, List<String> wordDict) {
+        Set<String> words = new HashSet<>(wordDict);
+        int n = s.length();
+        boolean[] dp = new boolean[n + 1];
         dp[0] = true;
-        for (int i = 1; i <= n; ++i)
-        {
-            for (int j = 0; j < i; ++j)
-            {
-                if (dp[j] && words.Contains(s.Substring(j, i - j)))
-                {
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 0; j < i; ++j) {
+                if (dp[j] && words.contains(s.substring(j, i))) {
                     dp[i] = true;
                     break;
                 }

solution/0100-0199/0139.Word Break/Solution.py

@@ -9,4 +9,4 @@ def wordBreak(self, s: str, wordDict: List[str]) -> bool:
                 if dp[j] and s[j:i] in words:
                     dp[i] = True
                     break
-        return dp[n]
+        return dp[-1]