LeetCode OJ

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".

解题思路：

　　方法一：利用深搜， 复杂度太高，导致超时

　　方法二：利用dp来做， dp[i][j] 表示s[i..j]可以被成功分割，则 dp[i][j] = dp[i][k] && dp[k+1][j] { i <= k <j}

　　方法三：同Word Break II , 利用DP来做。

此处采用方法二，复杂度为O（n^2)，方法三的空间复杂度会更低，代码如下：

　　

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        int n = s.length();
        vector<vector<bool>> in_dict(n, vector<bool>(n, false));// in_dict[j..i] means: s[j..i] can be beaked
        //init
        for (int i = 0; i < n; i++) {
            if (dict.count(s.substr(i, 1))) in_dict[i][i] = true;
        }

        for (int len = 2; len <= n; len++) {
            for (int start = 0; start <= n - len; start++) {
                string str = s.substr(start, len);
                if (dict.count(str)) in_dict[start][start + len - 1] = true;
                for (int i = start; i < start + len - 1; i++) {
                    if (in_dict[start][i] && in_dict[i + 1][start + len - 1]) {
                        in_dict[start][start + len - 1] = true;
                        break;
                    }
                }
            }
        }
        return in_dict[0][n - 1];
    }
};