139. Word Break

问题描述：

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

解题思路：

可以用dp来解。

对与s的每个位置，可以尝试向前匹配wordDict里面的每个单词。

即

　　　　　　　　　 int len = (int)w.size();
                if(i - len > -1 && dp[i-len]){
                    if(s.substr(i-len, len) == w){
                        dp[i] = true;
                        break;
                    }
                }

这里先检查了dp[i-len]是否能够匹配成功。

若能够，则检查这个长度的子串是否存在dict中

代码：

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int minLen = INT_MAX;
        int n = s.size();
        vector<bool> dp(n+1, false);
        dp[0] = true;
        for(int i = 0; i <= n; i++){
            for(auto w : wordDict){
                int len = (int)w.size();
                if(i - len > -1 && dp[i-len]){
                    if(s.substr(i-len, len) == w){
                        dp[i] = true;
                        break;
                    }
                }
            }
        }
        return dp[n];
    }
};