LeetCode 题解（87）: Word Break II
题目：
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"].
Word Break I + DFS， 用动态规划实现Word Break I，先回答s能否被wordDict里的字符串表示的问题。然后用递归实现DFS，找出所有可能的组合。如果不加Word Break I而直接DFS会在某些测试用例上超时，如s="aaaaaaaaab", wordDict = ["a", "ba"]
c++版：
class Solution {
public:
vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
vector<string> result;
if(s.length() == 0 || wordDict.size() == 0)
return result;
vector<bool> test(s.length()+1, false);
test[0] = true;
for(int i = 1; i <= s.length(); i++) {
for(int j = 0; j < i; j++) {
if(test[j] && wordDict.find(s.substr(j, i - j)) != wordDict.end()) {
test[i] = true;
break;
}
}
}
if(test[s.length()] == false)
return result;
if(wordDict.find(s) != wordDict.end())
result.push_back(s);
for(int i = 1; i < s.length(); i++) {
if(wordDict.find(s.substr(0, i)) != wordDict.end()) {
string front = s.substr(0, i);
vector<string> back = findRest(s.substr(i, s.length()-i), wordDict);
if(back.size() != 0) {
for(int j = 0; j < back.size(); j++) {
result.push_back(front + " " + back[j]);
}
}
}
}
return result;
}
vector<string> findRest(string s, unordered_set<string> &wordDict) {
vector<string> result;
if(wordDict.find(s) != wordDict.end())
result.push_back(s);
if(s.length() == 1)
return result;
for(int i = 1; i < s.length(); i++) {
if(wordDict.find(s.substr(0, i)) != wordDict.end()) {
string front = s.substr(0, i);
vector<string> back = findRest(s.substr(i, s.length()-i), wordDict);
if(back.size() != 0) {
for(int j = 0; j < back.size(); j++) {
result.push_back(front + " " + back[j]);
}
}
}
}
return result;
}
};
Java版：
public class Solution {
public List<String> wordBreak(String s, Set<String> wordDict) {
List<String> result = new ArrayList<>();
if(s == null || wordDict.isEmpty())
return result;
int[] test = new int[s.length()+1];
test[0] = 1;
for(int i = 1; i <= s.length(); i++) {
for(int j = 0; j < i; j++) {
if(test[j] == 1&& wordDict.contains(s.substring(j,i))) {
test[i] = 1;
}
}
}
if(test[s.length()] == 0)
return result;
if(wordDict.contains(s)) {
result.add(s);
}
for(int i = 1; i < s.length(); i++) {
if(wordDict.contains(s.substring(0, i))) {
String front = s.substring(0, i);
List<String> back = checkRest(s.substring(i), wordDict);
if(back.isEmpty() == false) {
for (String back1 : back) {
StringBuilder str = new StringBuilder();
str.append(front);
str.append(" ");
str.append(back1);
result.add(str.toString());
}
}
}
}
return result;
}
public List<String> checkRest(String s, Set<String> wordDict) {
List<String> result = new ArrayList<>();
if(s.length() == 1) {
if(wordDict.contains(s)) {
result.add(s);
}
return result;
}
if(wordDict.contains(s)) {
result.add(s);
}
for(int i = 1; i < s.length(); i++) {
if(wordDict.contains(s.substring(0,i))) {
String front = s.substring(0, i);
List<String> back = checkRest(s.substring(i), wordDict);
if(back.isEmpty() == false) {
for (String back1 : back) {
StringBuilder str = new StringBuilder();
str.append(front);
str.append(" ");
str.append(back1);
result.add(str.toString());
}
}
}
}
return result;
}
}
Python版：
class Solution:
# @param s, a string
# @param wordDict, a set<string>
# @return a string[]
def wordBreak(self, s, wordDict):
result = []
if len(s) == 0 or len(wordDict) == 0:
return result
test = [False] * (len(s) + 1)
test[0] = True
for i in range(1, len(s)+1):
for j in range(0, i):
if test[j] and s[j:i] in wordDict:
test[i] = True
break;
if test[len(s)] == False:
return result
if s in wordDict:
result.append(s)
for i in range(1, len(s)):
front = s[0:i]
if front in wordDict:
back = self.findRest(s[i:len(s)], wordDict)
if len(back) != 0:
for j in back:
result.append(front + " " + j)
return result
def findRest(self, s, wordDict):
result = []
if s in wordDict:
result.append(s)
if len(s) == 1:
return result
for i in range(1, len(s)):
front = s[0:i]
if front in wordDict:
back = self.findRest(s[i:len(s)], wordDict)
if len(back) != 0:
for j in back:
result.append(front + " " + j)
return result