LeetCode 139. Word Break

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

解析：给定一个非空的字符串和包含非空字符串的词典，判断字符串通过不同方式的切割，能够完全匹配词典中的词语。

采用动态规划进行求解，假设dp[i]表示以i结尾的字符串能否分割成词典中的词语。字符长度len=s.size()，则dp数组长度为len+1。外层循环控制每个下标i，内层循环控制j<i。假设dp[j]已经计算好，则判断从i到j的片段是否出现在词典中。如果出现了，并且dp[j]=true，则此时dp[i]=true。具体代码如下：

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