leetcode524- Longest Word in Dictionary through Deleting- medium

Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.

Example 1:

Input:
s = "abpcplea", d = ["ale","apple","monkey","plea"]

Output: 
"apple"

Example 2:

Input:
s = "abpcplea", d = ["a","b","c"]

Output: 
"a"

Note:

1. All the strings in the input will only contain lower-case letters.
2. The size of the dictionary won't exceed 1,000.
3. The length of all the strings in the input won't exceed 1,000.

算法。基本思路是遍历整个词典里的单词，拿着这个单词word去给入的新字符串s找每个word内字母有没有按顺序出现过。检查按顺序的方法有两种

1.用indexOf()每次更新。每次在后一点的位置开始找word的下一个字母。

2.(更简单)通过遍历s内的每个字母不断+1，同时一个指针指着word的字母偶尔+1，如果找到字符对上了，word的指针加上去，等到s的字母都走完了如果word的指针还没走到最后的话，就不合格。最后合格的word打个擂台就好。

时间复杂度：O(nk)。n是字符串长度，k是字典里单词个数。

细节：要求字典序比较方法是直接s1.compareTo(s2)

实现1：用indexOf的

class Solution {
    public String findLongestWord(String s, List<String> d) {
        String result = "";
        if (s == null || d == null || d.size() == 0) {
            return result;
        }

        for (String word : d) {
            int index = -1;
            int i = 0;
            for (i = 0; i < word.length(); i++) {
                index = s.indexOf(word.charAt(i), index);
                if (index == -1) {
                    break;
                }
                index++;
            }
            if (index != -1 && (word.length() > result.length() 
                                || word.length() == result.length() && word.compareTo(result) < 0)) {
                    result = word;
            }
        }
        return result;
    }
}

实现2：指针遍历

public String findLongestWord(String s, List<String> d) {
    String longest = "";
    for (String dictWord : d) {
        int i = 0;
        for (char c : s.toCharArray()) 
            if (i < dictWord.length() && c == dictWord.charAt(i)) i++;

        if (i == dictWord.length() && dictWord.length() >= longest.length()) 
            if (dictWord.length() > longest.length() || dictWord.compareTo(longest) < 0)
                longest = dictWord;
    }
    return longest;
}