LeetCode 139. Word Break单词拆分 (C++)

题目：

Given a non-empty string s and a dictionary wordDict containing a list of non-emptywords, 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

分析：

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict，判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词。

如果使用搜索的话应该是会超时的，这里使用动态规划的方法。

dp[i]表示字符串s的前i个字符能否被拆分，使用substr来截取字符串。

		l	e	e	t	c	o	d	e
dp	1	0	0	0	1	0	0	0	?

假如我们现在判断dp[i]的值，我们要同时判断dp[j]==1和后面的字符串(substr(j, i-j))是否在字典中是不是同时成立的。

比如现在要求dp[8]的值，j=0时，dp[0]==1但substr(0, 8-0)也就是leetcode不在字典中，当j=4时，dp[4]==1且substr(4, 8-4)也就是code在字典中，所以dp[8]赋值1。最后返回dp中最后的值即可。

		l	e	e	t	c	o	d	e
dp	1	0	0	0	1	0	0	0	1

 

注：在dp的长度要比字符串长度多1，是为了方便计算。

程序：

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