Longest Common Prefix 五种解法(JAVA)

Posted 2021-03-02 yawenw

解法一：水平扫描

int indexOf(String str): 在字符串中检索str，返回其第一出现的位置，如果找不到则返回-1

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if(strs.length == 0)
            return "";
        String prefix = strs[0];
        for(int i = 1; i < strs.length; i++){
            while(strs[i].indexOf(prefix) != 0){
                prefix = prefix.substring(0, prefix.length()-1);
                if(prefix.isEmpty())
                    return "";
            }
        }
        return prefix;
    }
}

 

解法二：垂直扫描

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if(strs.length == 0)
            return "";
        for(int i = 0; i < strs[0].length(); i++){
            char c = strs[0].charAt(i);
            for(int j = 1; j < strs.length; j++){
                if(strs[j].length() == i || strs[j].charAt(i) != c) //!!!关键点
                    return strs[0].substring(0, i);
            }
        }
        return strs[0];
    }
}

 

解法三：分治法

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if(strs.length == 0)
            return "";
        return subprefix(strs, 0, strs.length-1);
    }
    public String subprefix(String[] strs, int left, int right){
        if(left >= right)
            return strs[left];
        int mid = (left + right) / 2;
        String lsub = subprefix(strs, left, mid);
        String rsub = subprefix(strs, mid+1, right);//!!!分治法
        int min = Math.min(lsub.length(), rsub.length());
        for(int i = 0; i < min; i++){
            if(lsub.charAt(i) != rsub.charAt(i))
                return lsub.substring(0, i);
        }
        return lsub.substring(0, min);
    }
}

 

解法四：二分搜索法

public boolean startsWith(String prefix, int toffset)  prefix -- 前缀。 toffset -- 字符串中开始查找的位置

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if(strs.length == 0)
            return "";
        int minlen = Integer.MAX_VALUE;
        for(String str : strs)
            minlen = Math.min(minlen, str.length());
        int low = 1; //这里指第一个字符串
        int high = minlen;
        while(low <= high){
            int mid = (low+high)/2;
            if(isCommonPrefix(strs, mid)){
                low = mid+1;//+1
            }
            else{
                high = mid-1;//-1
            }
        }
        return strs[0].substring(0, (low+high)/2);
    }
    public boolean isCommonPrefix(String[] strs, int len){
        String str1 = strs[0].substring(0, len);
        for(int i = 1; i < strs.length; i++)
            if(!strs[i].startsWith(str1))
                return false;
        return true;
    }
}

 

解法五：Tire树

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 {

    // R links to node children
    private TrieNode[] links;

    private final int R = 26;

    private boolean isEnd;

    // number of children non null links
    private int size;    
    public void put(char ch, TrieNode node) {
        links[ch -‘a‘] = node;
        size++;
    }

    public int getLinks() {
        return size;
    }
    //assume methods containsKey, isEnd, get, put are implemented as it is described
   //in  https://leetcode.com/articles/implement-trie-prefix-tree/)
}

public class Trie {

    private TrieNode root;

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

//assume methods insert, search, searchPrefix are implemented as it is described
//in  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();
    }
}