139. 单词拆分

• 参考: https://leetcode-cn.com/problems/word-break/solution/dan-ci-chai-fen-by-leetcode/

• 暴力递归 超时，时间复杂度:O(N^N), 空间复杂度O(N): 递归的深度

    public boolean wordBreak(String s, List<String> wordDict) {
        return wordBreakSet(s, new HashSet<>(wordDict));
    }

    private boolean wordBreakSet(String s, HashSet<String> wordDict) {
        if (s.isEmpty()) {
            return true;
        }
        for (int i = 0; i < s.length(); i++) {
            String before = s.substring(0, i);
            String after = s.substring(i);
            if (wordDict.contains(after) && wordBreakSet(before, wordDict)) {
                return true;
            }
        }
        return false;
    }

• 递归+记忆 accept，时间复杂度:O(N^2), 空间复杂度O(N):递归的深度

public boolean wordBreak(String s, List<String> wordDict) {
        return wordBreakSet(s, new HashSet<>(wordDict), new HashSet<>());
    }

    private boolean wordBreakSet(String s, HashSet<String> wordDict, HashSet<String> memory) {
        if (s.isEmpty()) {
            return true;
        }
        if (memory.contains(s)) return false; //记忆
        for (int i = 0; i < s.length(); i++) {
            String before = s.substring(0, i);
            String after = s.substring(i);
            if (wordDict.contains(after) && wordBreakSet(before, wordDict, memory)) {
                return true;
            }
        }
        memory.add(s); //记忆
        return false;
    }

• 单纯的BFS，超时

public boolean wordBreak(String s, List<String> wordDict) {
        LinkedList<String> queue = new LinkedList<>();
        queue.add(s);
        while (!queue.isEmpty()) {
            String seg = queue.poll();
            for (String word : wordDict) {
                if (seg.startsWith(word)) {
                    String otherSeg = seg.substring(word.length());
                    if (otherSeg.isEmpty()) {
                        return true;
                    }
                    queue.add(otherSeg);
                }
            }
        }
        return false;
    }

• BFS + 记忆，accept

public boolean wordBreak(String s, List<String> wordDict) {
        LinkedList<String> queue = new LinkedList<>();
        queue.add(s);
        Set<String> memory = new HashSet<>(); //记忆
        while (!queue.isEmpty()) {
            String seg = queue.poll();
            if (!memory.contains(seg)) { //记忆
                for (String word : wordDict) {
                    if (seg.startsWith(word)) {
                        String otherSeg = seg.substring(word.length());
                        if (otherSeg.isEmpty()) {
                            return true;
                        }
                        queue.add(otherSeg);
                    }
                }
                memory.add(seg); //记忆
            }
        }
        return false;
    }

• dp 动态规划

public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] dp = new boolean[s.length()];//dp[i]代表字符串s截止到下标i时的子字符串是否可拆分，即 s.substring(0,i) 是否可拆分。
        dp[0] = true;
        Set<String> set = new HashSet<>(wordDict);
        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j >= 0; j--) {
                boolean before = dp[j];
                String middle = s.substring(j, i);
                String after = s.substring(i);
                if (before && (middle.isEmpty() || set.contains(middle)) && set.contains(after)) {
                    return true;
                }
                if (before && set.contains(middle)) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return false;
    }

• dp 动态规划(代码优化)

public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        Set<String> set = new HashSet<>(wordDict);
        for (int i = 0; i <= s.length(); i++) { //注意边界条件
            for (int j = 0; j < i; j++) {
                if (dp[j] && set.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
    }