给你一个字符串 s
和一个字符串列表 wordDict
作为字典。请你判断是否可以利用字典中出现的单词拼接出 s
。
注意:不要求字典中出现的单词全部都使用,并且字典中的单词可以重复使用。
示例 1:
输入: s = "leetcode", wordDict = ["leet", "code"] 输出: true 解释: 返回 true 因为 "leetcode" 可以由 "leet" 和 "code" 拼接成。
示例 2:
输入: s = "applepenapple", wordDict = ["apple", "pen"] 输出: true 解释: 返回 true 因为"
applepenapple"
可以由"
apple" "pen" "apple" 拼接成
。 注意,你可以重复使用字典中的单词。
示例 3:
输入: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"] 输出: false
提示:
1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s
和wordDict[i]
仅有小写英文字母组成wordDict
中的所有字符串 互不相同
动态规划法。
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。
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
words = set(wordDict)
n = len(s)
dp = [False] * (n + 1)
dp[0] = True
for i in range(1, n + 1):
for j in range(i):
if dp[j] and s[j:i] in words:
dp[i] = True
break
return dp[n]
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))) {
dp[i] = true;
break;
}
}
}
return dp[n];
}
}
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> words;
for (auto word : wordDict) {
words.insert(word);
}
int n = s.size();
vector<bool> dp(n + 1, false);
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()) {
dp[i] = true;
break;
}
}
}
return dp[n];
}
};
func wordBreak(s string, wordDict []string) bool {
words := make(map[string]bool)
for _, word := range wordDict {
words[word] = true
}
n := len(s)
dp := make([]bool, n+1)
dp[0] = true
for i := 1; i <= n; i++ {
for j := 0; j < i; j++ {
if dp[j] && words[s[j:i]] {
dp[i] = true
break
}
}
}
return dp[n]
}
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];
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)))
{
dp[i] = true;
break;
}
}
}
return dp[n];
}
}