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139. 单词拆分 #57

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@Geekhyt

Description

原题链接

状态定义

dp[i]: 字符串 s 的前 i 个字符子串 s[0..i-1] 是否由单词表组成。

状态转移方程

dp[i] = dp[j] && check(s[j..i-1])

  • 使用 j 对字符串进行分割,分割成 [0..j-1] (dp[j]) 和 [j, i-1]
  • check(s[j..i-1]):代表子串 s[j..i-1] 是否出现在字典中。

状态转移方程需要满足:

  1. dp[j] 为真
  2. 检查 s[j..i-1] 是否在单词表中

初始化

dp[0] = true,空串且合法。

const wordBreak = function(s, wordDict){
 const wordDictSet = new Set(wordDict)
 const len = s.length
 const dp = new Array(len + 1).fill(false)
 dp[0] = true
 for (let i = 1; i <= len; i++) {
 for (let j = i - 1; j >= 0; j--) {
 const suffix = s.slice(j, i)
 if (wordDictSet.has(suffix) && dp[j]) {
 dp[i] = true
 break
 }
 }
 }
 return dp[len]
}
  • 时间复杂度:O(n^2)
  • 空间复杂度:O(n)

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