学步园

The following solution is by shibai86

Bottom up DP

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        int len = s.length();
        vector<bool> dp(len + 1,false);
        dp[len] = true;

        for (int i = len - 1; i >= 0; i--) {
            for (int j = i; j < len; j++) {
                string str = s.substr(i,j - i + 1);
                if (dict.find(str) != dict.end() && dp[j + 1]) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[0];
    }
};

edit: dp[i] has the boolean value that means if string[i : n] can be partitioned according to our dictionary. Hence dp[len] represents the empty string and dp[0] is the answer we are looking for.

For each iteration in the inner for loop, we add a new cut at index j to string[i : n], in whcih string[i : j] has never checked before but string[j + 1 : n](result is in dp[j + 1]) has been known since the previous
iteration in outer loop.