给定一个有效的句子，在单词和有效的英语单词的词典之间没有任何空格，请找到所有可能的方法以将单词分解为单个词典的单词。
例子
Consider the following dictionary
{ i, like, sam, sung, samsung, mobile, ice,
and, cream, icecream, man, go, mango}
Input: "ilikesamsungmobile"
Output: i like sam sung mobile
i like samsung mobile
Input: "ilikeicecreamandmango"
Output: i like ice cream and man go
i like ice cream and mango
i like icecream and man go
i like icecream and mango我们在下面的文章中讨论了动态编程解决方案。
动态编程|设置32(分词问题)
动态编程解决方案仅查找断字是否可能。在这里，我们需要打印所有可能的分词符。
我们从左开始扫描句子。当我们找到有效的单词时，我们需要检查句子的其余部分是否可以生成有效的单词。因为在某些情况下，从左侧找到的第一个单词可能会留下无法进一步分离的剩余部分。因此，在这种情况下，我们应该返回并保留当前找到的单词，并继续搜索下一个单词。这个过程是递归的，因为要找出正确的部分是否可分离，我们需要相同的逻辑。因此，我们将使用递归和回溯来解决此问题。为了跟踪找到的单词，我们将使用堆栈。只要字符串的右边部分没有形成有效的单词，我们就会从堆栈中弹出顶部的字符串，然后继续查找。
以下是上述想法的实现：
CPP
// A recursive program to print all possible
// partitions of a given string into dictionary
// words
#include
using namespace std;
/* A utility function to check whether a word
is present in dictionary or not. An array of
strings is used for dictionary. Using array
of strings for dictionary is definitely not
a good idea. We have used for simplicity of
the program*/
int dictionaryContains(string &word)
{
string dictionary[] = {"mobile","samsung","sam","sung",
"man","mango", "icecream","and",
"go","i","love","ice","cream"};
int n = sizeof(dictionary)/sizeof(dictionary[0]);
for (int i = 0; i < n; i++)
if (dictionary[i].compare(word) == 0)
return true;
return false;
}
// Prototype of wordBreakUtil
void wordBreakUtil(string str, int size, string result);
// Prints all possible word breaks of given string
void wordBreak(string str)
{
// Last argument is prefix
wordBreakUtil(str, str.size(), "");
}
// Result store the current prefix with spaces
// between words
void wordBreakUtil(string str, int n, string result)
{
//Process all prefixes one by one
for (int i=1; i<=n; i++)
{
// Extract substring from 0 to i in prefix
string prefix = str.substr(0, i);
// If dictionary conatins this prefix, then
// we check for remaining string. Otherwise
// we ignore this prefix (there is no else for
// this if) and try next
if (dictionaryContains(prefix))
{
// If no more elements are there, print it
if (i == n)
{
// Add this element to previous prefix
result += prefix;
cout << result << endl;
return;
}
wordBreakUtil(str.substr(i, n-i), n-i,
result + prefix + " ");
}
}
}
//Driver Code
int main()
{
// Function call
cout << "First Test:\n";
wordBreak("iloveicecreamandmango");
cout << "\nSecond Test:\n";
wordBreak("ilovesamsungmobile");
return 0;
} Python3
# A recursive program to print all possible
# partitions of a given string into dictionary
# words
# A utility function to check whether a word
# is present in dictionary or not. An array of
# strings is used for dictionary. Using array
# of strings for dictionary is definitely not
# a good idea. We have used for simplicity of
# the program
def dictionaryContains(word):
dictionary = {"mobile", "samsung", "sam", "sung", "man",
"mango", "icecream", "and", "go", "i", "love", "ice", "cream"}
return word in dictionary
# Prints all possible word breaks of given string
def wordBreak(string):
# Last argument is prefix
wordBreakUtil(string, len(string), "")
# Result store the current prefix with spaces
# between words
def wordBreakUtil(string, n, result):
# Process all prefixes one by one
for i in range(1, n + 1):
# Extract substring from 0 to i in prefix
prefix = string[:i]
# If dictionary conatins this prefix, then
# we check for remaining string. Otherwise
# we ignore this prefix (there is no else for
# this if) and try next
if dictionaryContains(prefix):
# If no more elements are there, print it
if i == n:
# Add this element to previous prefix
result += prefix
print(result)
return
wordBreakUtil(string[i:], n - i, result+prefix+" ")
# Driver Code
if __name__ == "__main__":
print("First Test:")
wordBreak("iloveicecreamandmango")
print("\nSecond Test:")
wordBreak("ilovesamsungmobile")
# This code is contributed by harshitkap00r输出
First Test:
i love ice cream and man go
i love ice cream and mango
i love icecream and man go
i love icecream and mango
Second Test:
i love sam sung mobile
i love samsung mobile复杂性：
- 时间复杂度：O(n n )。因为在最坏的情况下有n n个组合。
- 辅助空间：O(n 2 )。由于在最坏的情况下wordBreakUtil(…)函数的递归堆栈。
其中n是输入字符串的长度。