LeetCode: Word Break

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".

Solution:

The key is not to use recursion，but iteration based on DP. This would save much computation. 

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<bool> checked(s.length()+1, false);
        checked[0] = true;
        for (int i = 1; i <= s.length(); ++i) {
            for (int j = 0; j < i; ++j) {
                if (checked[j] && dict.find(s.substr(j,i-j)) != dict.end()) {
                    checked[i] = true;
                    break;
                }
            }
        }
        return checked[s.length()];
    }
};