[leetcode] 139. Word Break

Description

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

Example 1:

Input:

s = "leetcode", wordDict = ["leet", "code"]

Output:

true

Explanation:

Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input:

s = "applepenapple", wordDict = ["apple", "pen"]

Output:

true

Explanation:

Return true because "applepenapple" can be segmented as "apple pen 
apple". Note that you are allowed to reuse a dictionary word.

Example 3:

Input:

 s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]

Output:

false

分析

题目的意思是：判断一个字符串能否分割成若干个字典中的单词。

• dp[i] 表示源串的前i个字符可以满足分割，那么 dp[ j ] 满足分割的条件是存在k 使得 dp [k] && substr[k,j]在字典里。

代码

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        vector<bool> dp(s.size()+1,false);
        dp[0]=true;
        for(int i=0;i<s.length();i++){
            for(int j=i;j<s.length()&&dp[i];j++){
                if(find(wordDict.begin(),wordDict.end(),s.substr(i,j-i+1))!=wordDict.end()){
                    dp[j+1]=true;
                }
            }
        }
        return dp[s.length()];
    }
};

参考文献

​​[编程题]word-break​​​​[Leetcode] Word Break、Word BreakII​​