给定一个非空序列S和一个包含非空单词列表的字典dict [] ，请打印所有可能的方法以将各个字典单词中的句子打断。
例子：
Input: S = “catsanddog”
dict[] = {“cat”, “cats”, “and”, “sand”, “dog”}
Output:
“cats and dog”
“cat sand dog”
Input: S = “pineapplepenapple”
dict[] = {“apple”, “pen”, “applepen”, “pine”, “pineapple”}
Output:
“pine apple pen apple”
“pineapple pen apple”
“pine applepen apple”
本文讨论了与此类似的问题，其中的任务是检查是否存在任何解决方案，以便可以将序列分解为词典单词。
方法：想法是检查从任何位置I开始的每个子字符串，使其结束于字典中存在的字符串的长度，然后简单地递归子字符串[0，I]。同时，为每个子字符串存储重叠的子问题，以避免再次计算子问题。重叠的子问题可以如下所示–
下面是上述方法的实现：
C++
// C++ implementation to break
// a sequence into the words of
// the dictionary
#include
using namespace std;
// Unordered_map used for storing
// the sentences the key string
// can be broken into
unordered_map > mp;
// Unordered_set used
// to store the dictionary.
unordered_set dict;
// Utility funtion used for
// printing the obtained result
void printResult(vector A)
{
for (int i = 0; i < A.size(); i++)
cout << A[i] << '\n';
}
// Utility function for
// appending new words
// to already existing strings
vector combine(
vector prev, string word){
// Loop to find the append string
// which can be broken into
for (int i = 0; i < prev.size(); i++) {
prev[i] += " " + word;
}
return prev;
}
// Utility funtion for word Break
vector wordBreakUtil(string s)
{
// Condition to check if the
// subproblem is already computed
if (mp.find(s) != mp.end())
return mp[s];
vector res;
// If the whole word is a dictionary
// word then directly append into
// the result array for the string
if (dict.find(s) != dict.end())
res.push_back(s);
// Loop to iterate over the substring
for (int i = 1; i < s.length(); i++) {
string word = s.substr(i);
// If the substring is present into
// the dictionary then recurse for
// other substring of the string
if (dict.find(word) != dict.end()) {
string rem = s.substr(0, i);
vector prev =
combine(wordBreakUtil(rem), word);
res.insert(res.end(),
prev.begin(), prev.end());
}
}
// Store the subproblem
// into the map
mp[s] = res;
return res;
}
// Master wordBreak function converts
// the string vector to unordered_set
vector wordBreak(string s,
vector& wordDict)
{
// Clear the previous stored data
mp.clear();
dict.clear();
dict.insert(wordDict.begin(), wordDict.end());
return wordBreakUtil(s);
}
// Driver Code
int main()
{
vector wordDict1 = {
"cat", "cats", "and", "sand", "dog" };
printResult(wordBreak("catsanddog", wordDict1));
return 0;
} Python3
# Python3 implementation to break
# a sequence into the words of
# the dictionary
# Unordered_map used for storing
# the sentences the key string
# can be broken into
mp = dict()
# Unordered_set used
# to store the dictionary.
dict_t = set()
# Utility funtion used for
# printing the obtained result
def printResult(A):
for i in range(len(A)):
print(A[i])
# Utility function for
# appending new words
# to already existing strings
def combine( prev, word):
# Loop to find the append string
# which can be broken into
for i in range(len(prev)):
prev[i] += " " + word;
return prev;
# Utility funtion for word Break
def wordBreakUtil(s):
# Condition to check if the
# subproblem is already computed
if (s in mp):
return mp[s];
res = []
# If the whole word is a dictionary
# word then directly append into
# the result array for the string
if (s in dict_t):
res.append(s);
# Loop to iterate over the substring
for i in range(1, len(s)):
word = s[i:];
# If the substring is present into
# the dictionary then recurse for
# other substring of the string
if (word in dict_t):
rem = s[:i]
prev = combine(wordBreakUtil(rem), word);
for i in prev:
res.append(i)
# Store the subproblem
# into the map
mp[s] = res;
return res;
# Master wordBreak function converts
# the string vector to unordered_set
def wordBreak(s, wordDict):
# Clear the previous stored data
mp.clear();
dict_t.clear();
for i in wordDict:
dict_t.add(i)
return wordBreakUtil(s);
# Driver Code
if __name__=='__main__':
wordDict1 = ["cat", "cats", "and", "sand", "dog" ]
printResult(wordBreak("catsanddog", wordDict1));
# This code is contributed by rutvik_56输出：
cat sand dog
cats and dog