LeetCode 14. Longest Common Prefix

最新推荐文章于 2024-01-22 08:26:35 发布

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最新推荐文章于 2024-01-22 08:26:35 发布

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分类专栏： LeetCode 文章标签： LeetCode Longest Common Prefix

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本文详细解析了LeetCode第14题“最长公共前缀”的多种解法，包括直接比较、垂直扫描等，提供了清晰的代码示例，帮助读者理解和掌握此题的解决策略。

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LeetCode 14. Longest Common Prefix

Description

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example

在这里插入图片描述

Note

All given inputs are in lowercase letters a-z.

Code

java

class Solution {
    public String longestCommonPrefix(String[] strs) {
        int len = strs.length;
        if(len == 0) return "";
        int commonLength = 0;
        int minLength = strs[0].length();
        for(int i = 1; i < len; i++) {
            minLength = Math.min(minLength, strs[i].length());
        }
        for(; commonLength < minLength; commonLength++) {
            char ch = strs[0].charAt(commonLength);
            boolean same = true;
            for(int i = 1; i < len; i++) {
                if(strs[i].charAt(commonLength) != ch) {
                    same = false;
                    break;
                }
            }
            if(!same) break;
        }
        return strs[0].substring(0, commonLength);
    }
}

Official solution1
按两个之间比较后得到的结果与下一个字符串继续比较。

 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;
}

Official Solution2：Vertical scanning

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs == null || 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 (i == strs[j].length() || strs[j].charAt(i) != c)
                    return strs[0].substring(0, i);             
            }
        }
        return strs[0];
    }
}