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.#include <string>
#include <vector>
#include <unordered_set>
using namespace std;
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> set;
int maxLen = 0;
vector<bool> flag(s.length() + 1, false);
for (const string& word : wordDict) {
set.insert(word);
maxLen = max(maxLen, (int)word.length());
}
for (int i = 1; i <= maxLen; ++i) {
if (dfs(s, 0, i, maxLen, set, flag)) {
return true;
}
}
return false;
}
private:
bool dfs(const string& s, int start, int end, int maxLen, unordered_set<string>& set, vector<bool>& flag) {
if (!flag[end] && set.count(s.substr(start, end - start))) {
flag[end] = true;
if (end == s.length()) {
return true;
}
for (int i = 1; i <= maxLen; ++i) {
if (end + i <= s.length() && dfs(s, end, end + i, maxLen, set, flag)) {
return true;
}
}
}
return false;
}
};