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

题解：

dp，使用set保存数据加快查找速度

class Solution {
public:
  bool wordBreak(string s, vector<string> &dict) {
    int n = s.length();
    int m = dict.size();
    unordered_set<string> st(dict.begin(), dict.end());
    if (n == 0 || m == 0) {
      return false;
    }
    if (st.find(s) != st.end()) {
      return true;
    }
    vector<bool> dp(n + 1, false);
    dp[0] = true;
    for (int i = 0; i < n; i++) {
      for (int j = 1; j <= n - i; j++) {
          if (dp[i + j] == true) {
              continue;
          }
          if (st.find(s.substr(i, j)) != st.end()) {
              dp[i + j] = dp[i];
          }
      }
    }
    return dp[n];
  }
};