leetcode14:最长公共前缀

编写一个函数来查找字符串数组中的最长公共前缀。

如果不存在公共前缀，返回空字符串 ""。

输入: ["flower","flow","flight"]
输出: "fl"

方法一：横向扫描

//Python3
class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        if len(strs) == 0:
            return ""
        return reduce(self.helper, strs)

    def helper(self, str1, str2):
        ans = ""
        if len(str1) == 0 or len(str2) == 0:
            return ans
        n = min(len(str1), len(str2))
        for i in range(n):
            if str1[i] == str2[i]:
                ans += str1[i]
            else:
                break
        return ans

//Java
class Solution {
    public String longestCommonPrefix(string[] strs) {
        if (strs == null || strs.length == 0) {
            return "";
        }
        String perfix = strs[0];
        int count = strs.length;
        for (int i = 1; i < count; i++) {
            prefix = longestCommonPrefix(prefix, strs[i]);
            if (prefix.length() == 0) {
                break;
            }
        }
        return prefix;
    }

    public String longestCommonPrefix(String str1, String str2) {
        int length = Math.min(str1.length(), str2.length());
        int index = 0;
        while (index < length && str1.charAt(index) == str2.charAt(index)) {
            index++;
        }
        return str1.substring(0, index);
    }
}

//Golang
func longestCommonPrefix(strs []string) string {
    if len(strs) == 0 {
        return ""
    }
    prefix := strs[0]
    count := len(strs)
    for i := 1; I < count; i++ {
        prefix = lcp(prefix, strs[I])
        if len(prefix) == 0 {
            break
        }
    }
    return prefix
}

func lcp(str1, str2 string) string {
    length := min(len(str1), len(str2))
    index := 0
    for index < length && str1[index] == str2[index] {
        index++
    }
    return str1[:index]
}

func min(x, y int) int {
    if x < y {
        return x
    }
    return y
}