解法参考：https://blog.csdn.net/yujin753/article/details/48010677
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
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict)
{
//边界条件 他说了不空 没有边界
//动态规划 一个个字符判断 必须前面是true在加上一个存在的字符串才可以
vector<bool> dp(s.size()+1,false);
int size = s.size();
dp[0] = true;
for (int i = 1; i <= size; i++)
{
for (int j = 0; j < i; j++)
{
if (dp[j] && find(wordDict.begin(), wordDict.end(), s.substr(j,i-j)) != wordDict.end())
{
dp[i] = true;
break;
}
}
}
//find(s.cbegin(),s.cend(),'1');
return dp[size];
}
};