Word Break
描述
Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s = "leetcode",
dict = ["leet", "code"].
Return true because "leetcode" can be segmented as "leet code".
分析
设状态为f(i)，表示s[0,i)是否可以分词，则状态转移方程为
f(i) = any_of(f(j) && s[j,i] in dict), 0 <= j < i
深搜
// Word Break
// 深搜，超时
// 时间复杂度O(2^n)，空间复杂度O(n)
class Solution {
public:
bool wordBreak(string s, unordered_set<string> &dict) {
return dfs(s, dict, 0, 1);
}
private:
static bool dfs(const string &s, unordered_set<string> &dict,
size_t start, size_t cur) {
if (cur == s.size()) {
return dict.find(s.substr(start, cur-start)) != dict.end();
}
if (dfs(s, dict, start, cur+1)) return true; // no cut
if (dict.find(s.substr(start, cur-start)) != dict.end()) // cut here
if (dfs(s, dict, cur+1, cur+1)) return true;
return false;
}
};
动规
// Word Break
// 动规，时间复杂度O(n^2)，空间复杂度O(n)
class Solution {
public:
bool wordBreak(string s, unordered_set<string> &dict) {
// 长度为n的字符串有n+1个隔板
vector<bool> f(s.size() + 1, false);
f[0] = true; // 空字符串
for (int i = 1; i <= s.size(); ++i) {
for (int j = i - 1; j >= 0; --j) {
if (f[j] && dict.find(s.substr(j, i - j)) != dict.end()) {
f[i] = true;
break;
}
}
}
return f[s.size()];
}
};