LeetCode-Word Break

LeetCode-Word Break

Given a stringSAnd a dictionary of wordsDict, Determine ifSCan be segmented into a space-separated sequence of one or more dictionary words.

For example, given
S="leetcode",
Dict=["leet", "code"].

Return true because"leetcode"Can be segmented"leet code".

In this example, we mainly want to see whether the words in a sentence string can be divided into words in the dictionary.

If flag [j] is true, a variable k must exist, which can satisfy the condition that flag [k] is true, and substr (k, j-k) is in the dictionary.

However, in this example, we do not care about the final division method.

 

Class Solution {public: bool wordBreak (string s, unordered_set
 
  
& Dict) {int length = s. size (); vector
  
   
Val (length + 1, false); val [0] = true; // <set int I = 0 according to the length; int j = 0; for (I = 0; I <length; I ++) {if (val [I]) // <the previous length can be decomposed {for (j = 1; (I + j) <= length; j ++) {string tmp = s. substr (I, j); if (dict. count (tmp)> 0) {val [I + j] = true ;}}} return val [length] ;}};
  
 

There is a second example:

 

Word Break II

 

 

Given a stringSAnd a dictionary of wordsDict, Add spaces inSTo 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"].

In this example, we need to put the decomposition result in a vector struct.

One method is the DFS method, which has not been studied yet.

The method in the first example is used as a reference here to check whether the partition can be made. If the partition can be made, it is then stored through traversal.

 

class Solution {public:    void breakWord(vector
 
   &res, string &s, unordered_set
  
    &dict, string str, int idx, vector
   
     &dp) {        string substr;        for (int len = 1; idx + len <= s.length(); ++len) {            if (dp[idx + len] && dict.count(s.substr(idx,len)) > 0) {                substr = s.substr(idx, len);                if (idx + len < s.length()) {                    breakWord(res, s, dict, str + substr + " ", idx + len, dp);                } else {                    res.push_back(str + substr);                    return;                }            }        }    }        vector
    
      wordBreak(string s, unordered_set
     
       &dict) {        vector
      
        dp(s.length() + 1, false); dp[0] = true; for (int i = 0; i < s.length(); ++i) { if (dp[i]) { for (int len = 1; i + len <= s.length(); ++len) { if (dict.count(s.substr(i, len)) > 0) { dp[i + len] = true; } } } } vector
       
         res; if (dp[s.length()]) breakWord(res, s, dict, "", 0, dp); return res; }};