Problem
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.
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 = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
"cats and dog",
"cat sand dog"
]
Example 2:
Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
"pine apple pen apple",
"pineapple pen apple",
"pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.
Example 3:
Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]
Solution
/**
* @param {string} s
* @param {string[]} wordDict
* @return {string[]}
*/
var wordBreak = function(s, wordDict) {
var dp = Array(s.length);
var map = {};
var res = [];
for (var i = 0; i < wordDict.length; i++) {
map[wordDict[i]] = true;
}
return find(s, map, dp, 0);
};
var find = function (s, map, dp, index) {
if (dp[index]) return dp[index];
var str = '';
var tmp = null;
var len = s.length;
dp[index] = [];
for (var i = index; i < len; i++) {
str = s.substring(index, i + 1);
if (!map[str]) continue;
if (i === len - 1) {
dp[index].push(str);
break;
}
tmp = find(s, map, dp, i + 1);
for (var j = 0; j < tmp.length; j++) {
dp[index].push(str + ' ' + tmp[j]);
}
}
return dp[index];
};
Explain:
dp[i] 代表 s.substring(i) 由 wordDict 里的词组成的方式。
Complexity:
- Time complexity : O(n^2).
- Space complexity : O(n^2).