Word Break

题目：

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. You may assume the dictionary does not contain duplicate words.

For example, given
s = “leetcode”,
dict = [“leet”, “code”].

Return true because “leetcode” can be segmented as “leet code”.

给定一个非空的字符串s以及一个字典wordDict ，wordDict 中包含一系列非空的单词。需要判定s是否能被分割为字典中的词。字典中不包含重复的单词。

思路：

这道题可以使用动态规划来解决。我们假设a[i]表示s[0:i]能够被分割为字典中的单词，那么如果s[j:i]在字典中，并且a[j]为True，那么自然a[i]也为True。如果i刚好等于s.length -1，那么我们就可以知道s能够被分割为字典中的单词。

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