LeetCode——最长公共前缀

Q：编写一个函数来查找字符串数组中的最长公共前缀。如果不存在公共前缀，返回空字符串 ""。

示例 1:
输入: ["flower","flow","flight"]
输出: "fl"

示例 2:
输入: ["dog","racecar","car"]
输出: ""
解释: 输入不存在公共前缀。

A：
最简单的方法：

public static String longestCommonPrefix(String[] strs) {
        if (strs.length == 0)
            return "";
        if (strs.length == 1)
            return strs[0];
        int minLen = Integer.MAX_VALUE;
        for (String s : strs) {
            int len = s.length();
            minLen = Math.min(minLen, len);
        }
        String sub = "";
        boolean flag = false;
        for (int i = 0; i < minLen; i++) {
            char c = strs[0].charAt(i);
            for (String str : strs) {
                if (str.charAt(i) != c) {
                    flag = true;
                    sub = str.substring(0, i);
                }
            }
            if (flag)
                break;
        }
        if (!flag) {
            sub = strs[0].substring(0, minLen);
        }
        return sub;
    }

对字符串数组进行排序，然后只要比较首尾两个字符串即可

    public String longestCommonPrefix(String[] strs) {
        StringBuilder result = new StringBuilder();
         
        if (strs!= null && strs.length > 0){
         
            Arrays.sort(strs);
             
            char [] a = strs[0].toCharArray();
            char [] b = strs[strs.length-1].toCharArray();
             
            for (int i = 0; i < a.length; i ++){
                if (b.length > i && b[i] == a[i]){
                    result.append(b[i]);
                }
                else {
                    return result.toString();
                }
            }
        return result.toString();
    }

分治法：

    public String longestCommonPrefix(String[] strs) {
        if (strs == null || strs.length == 0)
            return "";
        return longestCommon(strs, 0, strs.length - 1);
    }

    private String longestCommon(String[] strs, int start, int end) {
        if (start == end)
            return strs[start];
        int mid = (start + end) / 2;
        String left = longestCommon(strs, start, mid);
        String right = longestCommon(strs, mid + 1, end);
        return commonPrefix(left, right);
    }

    private String commonPrefix(String left, String right) {//这里也可以用二分查找
        int minLen = Math.min(left.length(), right.length());
        for (int i = 0; i < minLen; i++) {
            if (left.charAt(i) != right.charAt(i))
                return left.substring(0, i);
        }
        return left.substring(0, minLen);
    }

还有使用字典树（直接借用了别人的，链接：https://leetcode-cn.com/problems/longest-common-prefix/solution/zui-chang-gong-gong-qian-zhui-by-leetcode/）

public String longestCommonPrefix(String q, String[] strs) {
    if (strs == null || strs.length == 0)
         return "";  
    if (strs.length == 1)
         return strs[0];
    Trie trie = new Trie();      
    for (int i = 1; i < strs.length ; i++) {
        trie.insert(strs[i]);
    }
    return trie.searchLongestPrefix(q);
}

class TrieNode {

    // 子节点的链接数组
    private TrieNode[] links;

    private final int R = 26;

    private boolean isEnd;

    // 非空子节点的数量
    private int size;    
    public void put(char ch, TrieNode node) {
        links[ch -'a'] = node;
        size++;
    }

    public int getLinks() {
        return size;
    }
    // 假设方法 containsKey、isEnd、get、put 都已经实现了
    // 可以参考文章：https://leetcode.com/articles/implement-trie-prefix-tree/
}

public class Trie {

    private TrieNode root;

    public Trie() {
        root = new TrieNode();
    }

// 假设方法 insert、search、searchPrefix 都已经实现了
// 可以参考文章：https://leetcode.com/articles/implement-trie-prefix-tree/
    private String searchLongestPrefix(String word) {
        TrieNode node = root;
        StringBuilder prefix = new StringBuilder();
        for (int i = 0; i < word.length(); i++) {
            char curLetter = word.charAt(i);
            if (node.containsKey(curLetter) && (node.getLinks() == 1) && (!node.isEnd())) {
                prefix.append(curLetter);
                node = node.get(curLetter);
            }
            else
                return prefix.toString();

         }
         return prefix.toString();
    }
}