求最長前綴子串---indexOf()

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 1:

Input: ["flower","flow","flight"]
Output: "fl"
Example 2:

Input: ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.
Note:

All given inputs are in lowercase letters a-z.

思路一：採用java就是所有字符串一個個字符找過去，知道遇到不同的就輸出

class Solution {
    public String longestCommonPrefix(String[] strs) {
if(strs.length == 1)
            return strs[0];
        StringBuilder sb = new StringBuilder();
        
        if(strs.length > 1)
        {
            int len = strs[0].length();
            for(int i = 0; i < len; i++)
            {
                char currch = strs[0].charAt(i);
                for(int j = 1; j < strs.length; j++)
                {
                    
                   if(strs[j].length()<=i ||strs[j].charAt(i) != currch)
                        return sb.toString();
                    //當每個位置的值都相等，且來到最後一組了
                    if(strs[j].charAt(i) == currch && j == strs.length - 1)
                    {
                        sb.append(currch);
                    }
                }
            }
        }
        return sb.toString();
 }

思路二:採用indexOf()和subString,從第二個串開始找與第一個串的公共子串，然後跟第三個串也是同樣。就是空間複雜度有點高：

class Solution {
    public String longestCommonPrefix(String[] strs) {
       if(strs.length == 0)
            return "";
        String tmp = strs[0];
        for(int i = 1; i < strs.length; i++){
            //先跟第二個串匹配，找到相同的子串，每次沒找到公共子串，則
            //原始串則減1，就是最後一個，直到是公共串，然後匹配下一個子串
            while(strs[i].indexOf(tmp) != 0){
                tmp = tmp.substring(0, tmp.length()-1);
            }
        }
        return tmp;

    }
}