文章目录
problem
139. Word Break
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
wrong solution 1 Brute force recursive
class Solution {
public:
bool recur(int start, string s, unordered_set<string>& word){
if(start==s.size())return true;
string sub;
for(int i=start; i<s.size(); i++)
if(word.count(sub+=s[i]) && recur(i+1, s, word))
return true;
return false;
}
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> word(wordDict.begin(), wordDict.end());
return recur(0, s, word);
}
};
TLE
solution 2 dp
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
int n = s.size();
vector<bool> dp(n+1, false);
unordered_set<string> word(wordDict.begin(), wordDict.end());
dp[n]=true;
for(int i=n-1; i>=0; i--){
string sub;
for(int j=i; j<n; j++){
if(dp[i]=word.count(sub+=s[j]) && dp[j+1])
break;
}
}
return dp[0];
}
};
solution 3 DFS + prune
class Solution {
public:
bool recur(int start, string s, unordered_set<string>& word, vector<char>& mem){
int n = s.size();
if(start==n)return true;
if(mem[start] != -1)return mem[start];
string sub;
for(int i=start; i<n; i++)
if(word.count(sub+=s[i]) && recur(i+1, s, word, mem))
return mem[start]=1;
return mem[start]=0;
}
bool wordBreak(string s, vector<string>& wordDict) {
vector<char> mem(s.size(), -1);
unordered_set<string> word(wordDict.begin(), wordDict.end());
return recur(0, s, word, mem);
}
};