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Update 0139.单词拆分.md
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@ -369,8 +369,8 @@ class Solution:
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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
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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
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dp[0] = True # 初始状态,空字符串可以被拆分成单词
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dp[0] = True # 初始状态,空字符串可以被拆分成单词
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for i in range(1, n + 1):
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for i in range(1, n + 1): # 遍历背包
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for j in range(i):
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for j in range(i): # 遍历单词
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if dp[j] and s[j:i] in wordSet:
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if dp[j] and s[j:i] in wordSet:
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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
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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
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break
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break
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