Leetcode: Word Break

题目：
Given a string s and a dictionary of words dict, determine if s can 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 as “leet code”.

思路分析：
按照官方的解法：动态规划（Dynamic Programming）
感觉自己对动态规划问题还是不能熟练应对。不能准确找出变量之间的递推关系。
设F(0, i)表示前i个子串分割后是否存在于字典中。那么，这道题中存在这样的递推关系：F(0, i) = F(0, j) + F(j, i)，即前i个字符结果为true的充分条件是前j个字符结果为true，第j到i个子串结果也是true。

C++参考代码：

class Solution{public:    bool wordBreak(string s, unordered_set<string> &dict)    {        string::size_type size = s.size();        //申请size+1大小的空间，第一个设为true        vector<bool> matches(size + 1, false);        matches[0] = true;        for (size_t i = 1; i <= size; ++i)        {            for (size_t j = 0; j < i; ++j)            {                if (matches[j] && (dict.count(s.substr(j, i - j)) > 0))                {                    matches[i] = true;                    break;                }            }        }        return matches[size];    }};

C#参考代码：

public class Solution{    public bool WordBreak(string s, ISet<string> dict)    {        bool[] matches = new bool[s.Length + 1];        matches[0] = true;        for (int i = 1; i <= s.Length; ++i)        {            for (int j = 0; j < i; ++j)            {                if (matches[j] && dict.Contains(s.Substring(j, i - j)))                {                    matches[i] = true;                    break;                }            }        }        return matches[s.Length];    }}