Word Break II
描述
Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.
Return all such possible sentences.
For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].
A solution is ["cats and dog", "cat sand dog"].
分析
在上一题的基础上，要返回解本身。
代码
- Java
- C++
// Word Break II
// 动规，时间复杂度O(n^2)，空间复杂度O(n^2)
public class Solution {
public List<String> wordBreak(String s, Set<String> wordDict) {
// 长度为n的字符串有n+1个隔板
boolean[] f = new boolean[s.length() + 1];
// prev[i][j]为true，表示s[j, i)是一个合法单词，可以从j处切开
// 第一行未用
boolean[][] prev = new boolean[s.length() + 1][s.length()];
f[0] = true;
for (int i = 1; i <= s.length(); ++i) {
for (int j = i - 1; j >= 0; --j) {
if (f[j] && wordDict.contains(s.substring(j, i))) {
f[i] = true;
prev[i][j] = true;
}
}
}
List<String> result = new ArrayList<>();
List<String> path = new ArrayList<>();
gen_path(s, prev, s.length(), path, result);
return result;
}
// DFS遍历树，生成路径
private static void gen_path(String s, boolean[][] prev,
int cur, List<String> path, List<String> result) {
if (cur == 0) {
StringBuilder sb = new StringBuilder();
for (int i = path.size() - 1; i >= 0; --i)
sb.append(path.get(i)).append(' ');
sb.deleteCharAt(sb.length()-1);
result.add(sb.toString());
}
for (int i = 0; i < s.length(); ++i) {
if (prev[cur][i]) {
path.add(s.substring(i, cur));
gen_path(s, prev, i, path, result);
path.remove(path.size()-1);
}
}
}
}
// Word Break II
// 动规，时间复杂度O(n^2)，空间复杂度O(n^2)
class Solution {
public:
vector<string> wordBreak(string s, unordered_set<string> &dict) {
vector<bool> dp(s.length() + 1, false);
dp[0] = true;
// prev[i][j]为true，表示s[j, i)是一个合法单词，可以从j处切开
// 第一行未用
vector<vector<bool> > prev(s.length() + 1, vector<bool>(s.length()));
for (size_t i = 1; i <= s.length(); ++i) {
for (int j = i - 1; j >= 0; --j) {
if (dp[j] && dict.find(s.substr(j, i - j)) != dict.end()) {
dp[i] = true;
prev[i][j] = true;
}
}
}
vector<string> result;
vector<string> path;
gen_path(s, prev, s.length(), path, result);
return result;
}
private:
// DFS遍历树，生成路径
void gen_path(const string &s, const vector<vector<bool> > &prev,
int cur, vector<string> &path, vector<string> &result) {
if (cur == 0) {
string tmp;
for (auto iter = path.crbegin(); iter != path.crend(); ++iter)
tmp += *iter + " ";
tmp.erase(tmp.end() - 1);
result.push_back(tmp);
}
for (size_t i = 0; i < s.size(); ++i) {
if (prev[cur][i]) {
path.push_back(s.substr(i, cur - i));
gen_path(s, prev, i, path, result);
path.pop_back();
}
}
}
};