139 Word Break
139. Word Break
1. Question
Given a non-empty strings and a dictionary wordDict containing a list of non-empty words, determine if scan be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.
For example, given
s="leetcode"
,
dict=["leet", "code"]
.
Return true because"leetcode"
can be segmented as"leet code"
.
2. Implementation
(1) DFS + Memoinzation
class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
Map<Integer, Boolean> map = new HashMap<>();
Set<String> dict = new HashSet<>(wordDict);
return isBreakable(0, s, dict, map);
}
public boolean isBreakable(int start, String s, Set<String> dict, Map<Integer, Boolean> map) {
if (start == s.length()) {
return true;
}
if (map.containsKey(start)) {
return map.get(start);
}
boolean breakable = false;
for (int end = start + 1; end <= s.length(); end++) {
if (dict.contains(s.substring(start, end)) && isBreakable(end, s, dict, map)) {
breakable = true;
}
}
map.put(start, breakable);
return breakable;
}
}
(2) DP
class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
Set<String> dict = new HashSet<>(wordDict);
int n = s.length();
boolean[] dp = new boolean[n + 1];
dp[0] = true;
for (int i = 1; i <= n; i++) {
for (int j = 0; j < i; j++) {
if (dp[j] && dict.contains(s.substring(j, i))) {
dp[i] = true;
break;
}
}
}
return dp[n];
}
}
3. Time & Space Complexity
DFS + Memoinzation: 时间复杂度O(n^2), 空间复杂度O(n)
DP: 时间复杂度O(n^2), 空间复杂度O(n)
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