139. Word Break （DP)

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"

分析：注意超时问题

1. 动态规划问题，R[I]=R[0,J] && R[J,I] &&R[I,END].

public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] match = new boolean[s.length()+1];//注意长度比s大1，因为substring是左闭右开
        match[0] = true;//默认第一个是true才能开始循环
        for(int i=1;i<=s.length();i++){
            for(int j=0;j<i;j++){
                if(match[j] && wordDict.contains(s.substring(j,i))){
                    match[i] = true;
                    break;
                }
            }
        }
        return match[s.length()];//最后一个代表真个字符串的匹配情况
    }

2. 递归利用string 的startswith 方法

  /**
     * 搜索字符串是否可以被分割成单词串
     *
     * @param s       字符串
     * @param idx     处理的开始位置
     * @param wordMap 单词字典，开始字符相同的在同一个set集合中
     * @return 搜索结果
     */
    public boolean wordBreak(String s, int idx, Map<Character, Set<String>> wordMap) {

        if (idx >= s.length()) {
            return true;
        }

        Set<String> words = wordMap.get(s.charAt(idx));

        if (words != null) {
            for (String word : words) {
                // idx之前的字符已经匹配，如果从ide之后起匹配word单词
                if (s.startsWith(word, idx)) {
                    // 递归处理
                    boolean result = wordBreak(s, idx + word.length(), wordMap);
                    // 如果满足条件，返回true
                    if (result) {
                        return true;
                    }
                }
            }
        }

        return false;
    }

参考https://blog.csdn.net/DERRANTCM/article/details/47774547

分析：注意超时问题

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