题目
139. 单词拆分
官方题解
方法1:暴搜
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| class Solution {
public:
bool dfs(string s, vector<string>& wordDict, int u) {
if (u == s.size()) {
return true;
}
for (int i = 0; i < wordDict.size(); i ++) {
// 判断是否满足条件
bool flag = true;
for (int j = 0; j < wordDict[i].size(); j ++) {
if (u + j < s.size()) {
if (s[u + j] != wordDict[i][j]) {
cout << wordDict[i] << endl;;
flag = false;
break;
}
}
}
if (flag) {
if (dfs(s, wordDict, u + wordDict[i].size())) return true;;
}
}
return false;
}
bool wordBreak(string s, vector<string>& wordDict) {
return dfs(s, wordDict, 0);
}
};
|
时间复杂度:O(m^n),m为字典中单词的平均长度,n为字符串s的长度
空间复杂度:O(n)
方法2:动态规划
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| class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> wordset;
for (auto t : wordDict) {
wordset.insert(t);
}
vector<bool> dp(s.size() + 1);
dp[0] = true;
for (int i = 1; i <= s.size(); i ++) {
for (int j = 0; j < i; j ++) {
if (dp[j] && wordset.find(s.substr(j, i - j)) != wordset.end()) {
dp[i] = true;
break;
}
}
}
return dp[s.size()];
}
};
|
时间复杂度:O(n^2)
空间复杂度:O(n)