LeetCode——Word Break

LeetCode——Word Break

Question

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".

Solution

动态规划的思想，从头开始扫描字符串，判断当前子字符串是否可以在字典中查找到，取决于自身以及对这个子字符串的所有划分方式是否可以查找到。
递推关系式子： dp[i] = dp[j] && dp[i - j] (0 <= j <= i)
初始值 dp[0] = true; 表示子字符串长度为0的时候，是可以查找到的。

Answer

public:
    bool wordBreak(string s, vector<string>& wordDict) {
        if (wordDict.size() == 0 || s.empty())
            return false;
        vector<int> dp(s.length() + 1, false);
        dp[0] = true;
        for (int i = 1; i <= s.length(); i++) {
            for (int j = 0; j <= i; j++) {
                if (dp[j]) {
                	// 第二个参数表示，表示从j开始的字符个数
                    string str1 = s.substr((unsigned int)j, i - j);
                    if (check(wordDict, str1)) {
                        dp[i] = true;
                        break;
                    }
                }
            }
        }
        return dp[s.length()];
    }
    bool check(vector<string>& wordDict, string& str1) {
        for (string str : wordDict)
            if (str == str1)
                return true;
        return false;
    }
};