Problem

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

就是找出一個字串陣列中元素，最長的通用字首。例如：{“ab”,”abc”,”abd”}，答案是 “ab”。

Java 實現

package com.coderli.leetcode.algorithms.easy;

import java.util.Arrays;

/**
 * Write a function to find the longest common prefix string amongst an array of strings.
 *
 * @author li.hzh
 */
 

public class LongestCommonPrefix {

    public static void main(String[] args) {
        LongestCommonPrefix longestCommonPrefix = new LongestCommonPrefix();
        System.out.println(longestCommonPrefix.longestCommonPrefix(new String[]{"a"}));
        System.out.println(longestCommonPrefix.longestCommonPrefix
 
(new String[]{"a", "ab", "abc"}));
        System.out.println(longestCommonPrefix.longestCommonPrefix(new String[]{"abc", "ab", "abc"}));
        System.out.println(longestCommonPrefix.longestCommonPrefix(new String[]{"abd", "ab", "abc"}));
        System.out.println(longestCommonPrefix.longestCommonPrefix
 
(new String[]{"a", "b", "c"}));
    }

    public String longestCommonPrefix(String[] strs) {
        if (strs.length == 0) {
            return "";
        }
        if (strs.length == 1) {
            return strs[0];
        }
        String result = strs[0];
        for (int i = 1; i < strs.length; i++) {
            String currentStr = strs[i];
            int findLength = Math.min(result.length(), currentStr.length());
            int commonLength = 0;
            for (int j = 0; j < findLength; j++) {
                char charInResult = result.charAt(j);
                char charInCurrent = currentStr.charAt(j);
                if (charInResult == charInCurrent) {
                    commonLength++;
                } else {
                    break;
                }
            }
            result = result.substring(0,commonLength);
        }
        return result;
    }

    public String longestCommonPrefixBySortFirst(String[] strs) {
        if (strs.length == 0) {
            return "";
        }
        if (strs.length == 1) {
            return strs[0];
        }
        Arrays.sort(strs);
        String first = strs[0];
        String second = strs[strs.length - 1];
        String result = first;
        int commonLength = 0;
        for (int j = 0; j < first.length(); j++) {
            char charInFirst = result.charAt(j);
            char charInSecond = second.charAt(j);
            if (charInFirst == charInSecond) {
                commonLength++;
            } else {
                break;
            }
        }
        result = result.substring(0,commonLength);
        return result;
    }

}

分析

這裡提供了兩個解法。第一個解法longestCommonPrefix，應該最容易被想到的解法。就是直接暴力的遍歷陣列元素，拿每個元素跟當前結果做字首比較。找出最新的最大字首。輸出結果。需要注意處理一下邊界值就ok。

提供第二個解法longestCommonPrefixBySortFirst是因為，第一個解法測試雖然通過，但效率不高。因為考慮優化效能，減少遍歷。因此採用先排序再比較的做法。這樣只需要拿排序後的陣列的第一位和最後一位比較即可。減少一次遍歷，但是增加了一次排序損耗。不過快排一般來說是O（nlgn）的，而後一次的遍歷，遍歷的只是陣列最短長度元素的長度，因此提升了一些效率。

最後，我也檢視一下提交裡效率較高解法，主要的優化在於字首計算上。參考程式碼如下：

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;
}