題目鏈接
tag:

• Medium；
• DP；

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.

Note:
The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.

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

思路：
??DP解法的兩大難點(diǎn)，定義dp數(shù)組跟找出狀態(tài)轉(zhuǎn)移方程，先來(lái)看dp數(shù)組的定義，這里我們就用一個(gè)一維的dp數(shù)組，其中dp[i]表示范圍[0, i)內(nèi)的子串是否可以拆分，注意這里dp數(shù)組的長(zhǎng)度比s串的長(zhǎng)度大1，是因?yàn)槲覀円幚砜沾那闆r，我們初始化dp[0]為true，然后開(kāi)始遍歷。注意這里我們需要兩個(gè)for循環(huán)來(lái)遍歷，我們必須要遍歷所有的子串，我們用j把[0, i)范圍內(nèi)的子串分為了兩部分，[0, j) 和 [j, i)，其中范圍 [0, j) 就是dp[j]，范圍 [j, i) 就是s.substr(j, i-j)，其中dp[j]是之前的狀態(tài)，已經(jīng)算出來(lái)了，可以直接用，只需要在字典中查找s.substr(j, i-j)是否存在了，如果二者均為true，將dp[i]賦為true，并且break掉，此時(shí)就不需要再用j去分[0, i)范圍了，因?yàn)閇0, i)范圍已經(jīng)可以拆分了。最終我們返回dp數(shù)組的最后一個(gè)值，代碼如下：

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        // DP
        unordered_set<string> wordSet(wordDict.begin(), wordDict.end());
        vector<bool> dp(s.size() + 1);
        dp[0] = true;
        for (int i = 0; i < dp.size(); ++i) {
            for (int j = 0; j < i; ++j) {
                if (dp[j] && wordSet.count(s.substr(j, i - j))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp.back();
    }
};