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

---

给定一个词典和一个字符串，问字符串能否拆分成若干个词典中有的词。比如样例中词典中有leet和code，则给定的字符串leetcode就能拆分成leet和code，他们都在词典中。
使用DP思路。假设dp[j]表示字符串下标j之前的字符串能否拆分成词典中的词的组合。如果要求dp[i]，则我们可以看看i之前的所有j，如果有dp[j]是true的，且[j,i)的子串在dict中，说明[0,i)也能被拆分成词典中的词，所以dp[i]=true。最后我们只需要返回dp[n]即可。
对于s=”leetcode”，dp=1,0,0,0,1,0,0,0,1，所以dp[n]=1。
代码如下：

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

本代码提交AC，用时3MS。