139. Word Break(Leetcode每日一題-2020.06.25）

Problem

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.

Example1

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

Example2

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.

Example3

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

Solution

動態規劃

狀態表示

轉移方程

用指針 j去劃分兩部分
[0, i] 區間子串 的 dp[i+1] 爲真，取決於兩部分：
它的前綴子串 [0, j-1] 的 dp[j] 爲真
剩餘子串 [j,i] 是一個合格的單詞

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int n = s.size();
        if(n == 0)
            return true;
        vector<bool> dp(n+1,false);
        unordered_set<string> wordDictSet(wordDict.begin(),wordDict.end());
        dp[0] = true;
        for(int i =1;i<=n;i++)
        {
            if(dp[i-1])
            {
                int indx = i-1;
                for(int j = indx;j<n;j++)
                {
                    string cur = s.substr(indx,j-indx+1);
                    if(wordDictSet.find(cur) != wordDictSet.end())
                        dp[j+1] = true;
                }
            }
        }
        return dp[n];

    }
};

Ref

https://leetcode-cn.com/problems/word-break/solution/shou-hui-tu-jie-san-chong-fang-fa-dfs-bfs-dong-tai/