LeetCode 139. Word Break（单词分隔）

原题网址：https://leetcode.com/problems/word-break/

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

方法一：递归+记忆化搜索。

public class Solution {    private boolean[] visited;    private boolean[] canBreak;    private boolean dp(String s, int to, Set<String> wordDict) {        if (to == 0) return true;        if (visited[to-1]) return canBreak[to-1];        for(int i=0; i<to; i++) {            if (wordDict.contains(s.substring(i)) && dp(s.substring(0,i), i, wordDict)) {                visited[to-1] = true;                canBreak[to-1] = true;                return true;            }        }        visited[to-1] = true;        canBreak[to-1] = false;        return false;    }    public boolean wordBreak(String s, Set<String> wordDict) {        visited = new boolean[s.length()];        canBreak = new boolean[s.length()];        return dp(s, s.length(), wordDict);    }}

方法二：动态规划。

public class Solution {    public boolean wordBreak(String s, Set<String> wordDict) {        if (null == s || "".equals(s)) return true;        boolean[] canBreak = new boolean[s.length()];        canBreak[0] = wordDict.contains(s.substring(0,1));        for(int i=1; i<s.length(); i++) {            if (wordDict.contains(s.substring(0, i+1))) {                canBreak[i] = true;            } else {                for(int j=0; j<i; j++) {                    if (canBreak[j] && wordDict.contains(s.substring(j+1, i+1))) {                        canBreak[i] = true;                        break;                    }                }            }        }        return canBreak[canBreak.length-1];    }}