Medium
Given a string s
and a dictionary of strings wordDict
, return true
if s
can be segmented into a space-separated sequence of one or more dictionary words.
Note that the same word in the dictionary may be reused multiple times in the segmentation.
Example 1:
Input: s = “leetcode”, wordDict = [“leet”,”code”]
Output: true
Explanation: Return true because “leetcode” can be segmented as “leet code”.
Example 2:
Input: s = “applepenapple”, wordDict = [“apple”,”pen”]
Output: true
Explanation: Return true because “applepenapple” can be segmented as “apple pen apple”.
Note that you are allowed to reuse a dictionary word.
Example 3:
Input: s = “catsandog”, wordDict = [“cats”,”dog”,”sand”,”and”,”cat”]
Output: false
Constraints:
1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s
and wordDict[i]
consist of only lowercase English letters.wordDict
are unique.class Solution {
bool wordBreak(String s, List<String> wordDict) {
Map<String, bool> memo = {};
bool solve(String s1) {
if (s1.length == 0) return true;
if (memo.containsKey(s1)) {
return memo[s1]!;
}
bool ans = false;
for (var word in wordDict) {
if (word.length > s1.length) continue;
String s2 = s1.substring(0, word.length);
if (s2 == word) {
ans |= solve(s1.substring(word.length));
}
}
memo[s1] = ans;
return ans;
}
return solve(s);
}
}