139. 单词拆分

原题链接：

139. 单词拆分

https://leetcode.cn/problems/word-break/description/

完成情况：

解题思路：

参考代码：

_139单词拆分01

package 代码随想录.动态规划;

import java.util.HashSet;
import java.util.List;

public class _139单词拆分01 {
    /**
     *
     * @param s 字符串 s
     * @param wordDict  字符串列表 wordDict 作为字典
     * @return  请你判断是否可以利用字典中出现的单词拼接出 s 。
     */
    public boolean wordBreak(String s, List<String> wordDict) {
        HashSet<String> set = new HashSet<String>(wordDict);
        boolean[] valid = new boolean[s.length() + 1];
        valid[0] = true;
        for (int i=1;i<= s.length();i++){
            for (int j=0;j<i && !valid[i];j++){
                if (set.contains(s.substring(j,i)) && valid[j]){
                    valid[i] = true;
                }
            }
        }
        return valid[s.length()];
    }
}

_139单词拆分02

package 代码随想录.动态规划;

import java.util.List;

public class _139单词拆分02 {
    /**
     *
     * @param s
     * @param wordDict
     * @return
     */
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean [] dp =  new boolean[s.length() + 1];
        dp[0] = true;

        for (int i=1;i<=s.length();i++) {
            for (String word : wordDict) {
                int len = word.length();
                if (i>=len && dp[i - len] && word.equals(s.substring(i-len,i))){
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
    }
}

_139单词拆分_回溯法and记忆法

package 代码随想录.动态规划;

import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class _139单词拆分_回溯法and记忆法 {
    private Set<String> sets;
    private int [] memo;
    /**
     *
     * @param s
     * @param wordDict
     * @return
     */
    public boolean wordBreak(String s, List<String> wordDict) {
        memo = new int[s.length()];
        sets = new HashSet<String>(wordDict);
        return backTrack(s,0);
    }

    /**
     *
     * @param s
     * @param startIndex
     * @return
     */
    private boolean backTrack(String s, int startIndex) {
        if (startIndex == s.length()){
            return true;
        }
        if (memo[startIndex] == -1){
            return false;
        }
        for (int i=startIndex;i<s.length();i++){
            String subsets = s.substring(startIndex,i+1);
            //拆分出来的单词无法匹配
            if (!sets.contains(subsets)){
                continue;
            }
            boolean res = backTrack(s,i+1);
            if (res){
                return true;
            }
        }
        //这里是关键，找遍了startIndex~~s.length()也没能完全匹配
        memo[startIndex] = -1;
        return false;
    }
}

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