139. Word Break

题目：

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

Input: s = "leetcode", wordDict = ["leet","code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple","pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output: false

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

思路：

一道完全背包题，dp[j]表示s中的j个字母形成的字符串是否能够被字典内的单词所形成。这里转换过来以后，字符串长度就是背包的重量，而单词就是每个物品，只不过物品的价值是能够构成字符串，因此是布尔值。转移公式是dp[j] = dp[j] || dp[j - len]，这里len是当前字典中的单词的长度，转移公式的含义即如果从j - len位置到j位置正好是当前单词wordDict[i]的话，则只要dp[j - len]为真则当前dp[j]即可为真。

代码：

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int n = s.size();
        vector<bool> f(n + 1, false);
        f[0] = true;
        for (int j = 1; j < n + 1; j++) {
            for (int i = 0; i < wordDict.size(); i++) {
                int len = wordDict[i].size();
                if (j < len)
                    continue;
                bool same = true;
                for (int k = 0; k < len; k++) {
                    if (s[j - (len - k)] != wordDict[i][k]) {
                        same = false;
                        break;
                    }
                }
                if (same) {
                    f[j] = f[j] || f[j - len];
                }
            }
        }
        return f[n];
    }
};