题目链接
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解法的两大难点，定义dp数组跟找出状态转移方程，先来看dp数组的定义，这里我们就用一个一维的dp数组，其中dp[i]表示范围[0, i)内的子串是否可以拆分，注意这里dp数组的长度比s串的长度大1，是因为我们要处理空串的情况，我们初始化dp[0]为true，然后开始遍历。注意这里我们需要两个for循环来遍历，我们必须要遍历所有的子串，我们用j把[0, i)范围内的子串分为了两部分，[0, j) 和 [j, i)，其中范围 [0, j) 就是dp[j]，范围 [j, i) 就是s.substr(j, i-j)，其中dp[j]是之前的状态，已经算出来了，可以直接用，只需要在字典中查找s.substr(j, i-j)是否存在了，如果二者均为true，将dp[i]赋为true，并且break掉，此时就不需要再用j去分[0, i)范围了，因为[0, i)范围已经可以拆分了。最终我们返回dp数组的最后一个值，代码如下：
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();
}
};
