From d8c51b2d362de5a72ec86a7df3f31e486aa8febf Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Mon, 5 Jun 2023 03:09:10 -0500
Subject: [PATCH] =?UTF-8?q?Update=200139.=E5=8D=95=E8=AF=8D=E6=8B=86?=
 =?UTF-8?q?=E5=88=86.md?=
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---
 problems/0139.单词拆分.md | 56 +++++++++++++++++++++++++++--------
 1 file changed, 44 insertions(+), 12 deletions(-)

diff --git a/problems/0139.单词拆分.md b/problems/0139.单词拆分.md
index 230942ef..0f24c71e 100644
--- a/problems/0139.单词拆分.md
+++ b/problems/0139.单词拆分.md
@@ -337,10 +337,53 @@ class Solution {
 
 Python：
 
+回溯
+```python
+class Solution:
+    def backtracking(self, s: str, wordSet: set[str], startIndex: int) -> bool:
+        # 边界情况：已经遍历到字符串末尾，返回True
+        if startIndex >= len(s):
+            return True
+
+        # 遍历所有可能的拆分位置
+        for i in range(startIndex, len(s)):
+            word = s[startIndex:i + 1]  # 截取子串
+            if word in wordSet and self.backtracking(s, wordSet, i + 1):
+                # 如果截取的子串在字典中，并且后续部分也可以被拆分成单词，返回True
+                return True
+
+        # 无法进行有效拆分，返回False
+        return False
+
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        wordSet = set(wordDict)  # 转换为哈希集合，提高查找效率
+        return self.backtracking(s, wordSet, 0)
+
+```
+DP（版本一）
+```python
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        wordSet = set(wordDict)
+        n = len(s)
+        dp = [False] * (n + 1)  # dp[i] 表示字符串的前 i 个字符是否可以被拆分成单词
+        dp[0] = True  # 初始状态，空字符串可以被拆分成单词
+
+        for i in range(1, n + 1):
+            for j in range(i):
+                if dp[j] and s[j:i] in wordSet:
+                    dp[i] = True  # 如果 s[0:j] 可以被拆分成单词，并且 s[j:i] 在单词集合中存在，则 s[0:i] 可以被拆分成单词
+                    break
+
+        return dp[n]
+
+
+```
+DP（版本二）
+
 ```python
 class Solution:
     def wordBreak(self, s: str, wordDict: List[str]) -> bool:
-        '''排列'''
         dp = [False]*(len(s) + 1)
         dp[0] = True
         # 遍历背包
@@ -351,17 +394,6 @@ class Solution:
                     dp[j] = dp[j] or (dp[j - len(word)] and word == s[j - len(word):j])
         return dp[len(s)]
 ```
-```python
-class Solution: # 和视频中写法一致（和最上面C++写法一致）
-    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
-        dp = [False]*(len(s)+1)
-        dp[0]=True
-        for j in range(1,len(s)+1):
-            for i in range(j):
-                word = s[i:j]
-                if word in wordDict and dp[i]: dp[j]=True
-        return dp[-1]
-```