LeetCode: Longest Common Prefix 解题报告-优快云博客

本文深入探讨了在字符串数组中寻找最长公共前缀的两种高效算法实现，通过详细的代码示例，阐述了如何遍历字符串并比较字符，以确定共同的前缀。文章提供了清晰的边界条件处理思路，避免了常见的错误陷阱。

Longest Common Prefix Total Accepted: 24665 Total Submissions: 92370

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Question

Solution

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

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SOULTION1:

解法就是扫一次。不过各种边界条件很容易出错。

 1 public class Solution {
 2     public String longestCommonPrefix(String[] strs) {
 3         if (strs == null) {
 4             return null;
 5         }
 6         
 7         if (strs.length == 0 || strs[0].length() == 0) {
 8             return "";
 9         }
10         
11         int len = strs[0].length();
12         int i = 0;
13         for (; i < len; i++) {
14             char c = strs[0].charAt(i);
15             
16             int j = 1;
17             for (; j < strs.length; j++) {
18                 if (strs[j].length() <= i || c != strs[j].charAt(i)) {
19                     break;
20                 }
21             }
22             
23             // there is err.
24             if (j < strs.length) {
25                 break;
26             }
27         }
28         
29         // The char i is invalid. 因为读到i时退出，所以不应包含i本身。
30         return strs[0].substring(0, i);
31     }
32 }

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SOULTION2:

感谢大神的灵感http://blog.youkuaiyun.com/fightforyourdream/article/details/14642079

重新改写：

 1 public class Solution {
 2     //http://blog.youkuaiyun.com/fightforyourdream/article/details/14642079
 3     public String longestCommonPrefix(String[] strs) {
 4         if (strs == null) {
 5             return null;
 6         }
 7         
 8         if (strs.length == 0) {
 9             return "";
10         }
11         
12         String s = strs[0];
13         int len = s.length();
14         
15         for (int i = 0; i < len; i++) {
16             char c = s.charAt(i);
17             
18             for (int j = 1; j < strs.length; j++) {
19                 if (strs[j].length() <= i || c != strs[j].charAt(i)) {
20                     // The char i is invalid. 因为读到i时退出，所以不应包含i本身。
21                     return s.substring(0, i);
22                 }
23             }
24         }
25         
26         // Didn't break, the whole String is valid.
27         return s;
28     }
29 }