LeetCode 139. Word Break

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".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.
 

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

分析題意可以發現，最終的答案可以由多個不同的路徑走過來，但是題目要求返回的只是最終能不能，因此，我們只需要在每個點返回可以不可以這樣一個boolean就行，另外再用一個boolean陣列記錄是否已經走過此位置，完成優化。但是iteration在理論上是沒有進行優化的，就是全部都走完，只有一個剪枝優化就是可以提前返回，但是iteration在空間和語法上是有優化的，因為陣列可以bulk load，但是recursion需要不停的從硬碟上來回取資料。

 //iteration版本
    public boolean wordBreak(String s, List<String> wordDict) {
        HashSet<String> set = new HashSet<>();
        for (String ss : wordDict) {
            set.add(ss);
        }
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                if (set.contains(s.substring(i,j+1)) && dp[i]) {
                    if (j == s.length() - 1) return true;
                    dp[j+1] = true;
                }
            }
        }
        return false;
    }
    // recursion版本
    public boolean wordBreak(String s, List<String> wordDict) {
        HashSet<String> set = new HashSet<>();
        for (String ss : wordDict) {
            set.add(ss);
        }
        boolean[] dp = new boolean[s.length()];
        return recursion(s, set, 0, dp);
    }
    private boolean recursion(String s, HashSet<String> set, int index, boolean[] dp) {
        //base case
        if (index == s.length()) return true;
        if (dp[index] == true) return false;
        //recursion
        boolean b = false;
        for (int i = s.length() - 1; i >= index; i--) {
            if (set.contains(s.substring(index, i+1))) {
                b = b || recursion(s, set, i+1, dp);
            }
        }
        dp[index] = true;
        return b;
    }