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
题意：判断给出的字符串是否可以被 wordDict 中的字符串拼凑出来。
思路：动态规划的典型题目。
代码：
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> wd(wordDict.begin(), wordDict.end());
vector<bool> dp(s.size() + 1, false);
//dp[i]表示s的前i个字符可以被wordDict拼凑出来
//dp[0]即s的前0个字符"", 为true
dp[0] = true;
//每次循环, 检查长度为i的字符串区间是否可以被拼凑出来,直到检查整个s
for (int i = 1; i <= s.size(); ++i) {
//每次循环, 若前面长j的字符串可以被拼出来, 则检查从j开始的长i区间
//的后一部分长为i-j的字符串是否可以在wordDict中找到
//接着依次检查长为j+1,j+2,...,i-1的字符串是否可以被拼出来,...
for (int j = 0; j < i; ++j) {
if (dp[j] && wd.find(s.substr(j, i - j)) != wd.end()) {
//前面的长j字符串和后面长i-j的字符串均可拼出
//则s的前i个字符都可拼凑, dp[i]为true
dp[i] = true;
break;
}
}
}
return dp[s.size()];
}
};