139. Word Break

Problem

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

Solution (Java)

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> set = new HashSet<>();
        int max = 0;
        boolean[] flag = new boolean[s.length() + 1];
        for (String st : wordDict) {
            set.add(st);
            if (max < st.length()) {
                max = st.length();
            }
        }
        for (int i = 1; i <= max; i++) {
            if (dfs(s, 0, i, max, set, flag)) {
                return true;
            }
        }
        return false;
    }

    private boolean dfs(String s, int start, int end, int max, Set<String> set, boolean[] flag) {
        if (!flag[end] && set.contains(s.substring(start, end))) {
            flag[end] = true;
            if (end == s.length()) {
                return true;
            }
            for (int i = 1; i <= max; i++) {
                if (end + i <= s.length() && dfs(s, end, end + i, max, set, flag)) {
                    return true;
                }
            }
        }
        return false;
    }
}

Solution (Javascript)

/**
 * @param {string} s
 * @param {string[]} wordDict
 * @return {boolean}
 */
var wordBreak = function(s, wordDict) {
  var len = wordDict.length;
  var dp = Array(len);
  var map = {};
  for (var i = 0; i < len; i++) {
    map[wordDict[i]] = true;
  }
  return find(s, map, dp, 0);
};

var find = function (s, map, dp, index) {
  if (dp[index] !== undefined) return dp[index];

  var str = '';
  var res = false;
  var len = s.length;

  if (index === len) return true;

  for (var i = index; i < len; i++) {
    str = s.substring(index, i + 1);
    if (map[str] && find(s, map, dp, i + 1)) {
      res = true;
      break;
    }
  }

  dp[index] = res;
  return res;
};

Explain:

dp[i] 代表 s.substring(i) 是否可以由 wordDict 里的词组成。

Complexity:

• Time complexity : O(n^2).
• Space complexity : O(n^2).