在输入此问题时,给出的一个句子不带空格,还提供了另一本词典,其中包含一些有效的英语单词。我们必须找到在单个词典单词中打断句子的可能方法。
当找到有效单词时,我们将尝试从字符串的左侧搜索以找到有效单词,我们将在该字符串的下一部分中搜索单词。
Input: A set of valid words as dictionary, and a string where different words are placed without spaces. Dictionary: {mobile, sam, sung, man, mango, icecream, and, go, i, love, ice, cream} Given String: “ilovemangoicecream” Output: All possible ways to break the string into given words. i love man go ice cream i love man go icecream i love mango ice cream i love mango icecream
wordBreak(string, n, result)
输入-给定的字符串,字符串的长度,分隔的字符串。
输出-使用字典分隔字符串。
Begin for i := 0 to n, do subStr := substring of given string from (0..i) if subStr is in dictionary, then if i = n, then result := result + subStr display the result return wordBreak(substring from (i..n-i), n-i, result, subStr, ‘space’) done End
#include <iostream> #define N 13 using namespace std; string dictionary[N] = {"mobile","samsung","sam","sung","man","mango", "icecream","and", "go","i","love","ice","cream"}; int isInDict(string word){ //check whether the word is in dictionary or not for (int i = 0; i < N; i++) if (dictionary[i].compare(word) == 0) return true; return false; } void wordBreak(string str, int n, string result) { for (int i=1; i<=n; i++) { string subStr = str.substr(0, i); //get string from 0 to ith location of the string if (isInDict(subStr)) { //if subStr is found in the dictionary if (i == n) { result += subStr; //add substring in the result. cout << result << endl; return; } wordBreak(str.substr(i, n-i), n-i, result + subStr + " "); //otherwise break rest part } } } int main() { string str="iloveicecreamandmango"; wordBreak(str, str.size(),""); }
输出结果
i love man go ice cream i love man go icecream i love mango ice cream i love mango icecream