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. 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".

比如这题，可以用一个数组dp记录S[0:i]是否能break。

dp[i]：S[0:i-1]是否能被break。例如dp[1]表示s[0]是否能被break。 dp[0] = true dp[i] = true if and only if:

1. 存在一个i-1>=k>=0，使s[k:i-1]存在于dict中。
2. s[0: k-1]可以被break，即dp[k] = true。

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

REF:

• http://bangbingsyb.blogspot.ca/2014/11/leetcode-word-break-i-ii.html
• http://blog.csdn.net/linhuanmars/article/details/22358863
• http://www.cnblogs.com/grandyang/p/4257740.html