Skip to content

Latest commit

 

History

History
134 lines (101 loc) · 2.75 KB

File metadata and controls

134 lines (101 loc) · 2.75 KB

English Version

题目描述

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 ""

 

示例 1:

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

示例 2:

输入:strs = ["dog","racecar","car"]
输出:""
解释:输入不存在公共前缀。

 

提示:

  • 0 <= strs.length <= 200
  • 0 <= strs[i].length <= 200
  • strs[i] 仅由小写英文字母组成

解法

Python3

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        n = len(strs)
        if n == 0:
            return ''
        for i in range(len(strs[0])):
            for j in range(1, n):
                if len(strs[j]) <= i or strs[j][i] != strs[0][i]:
                    return strs[0][:i]
        return strs[0]

Java

class Solution {
    public String longestCommonPrefix(String[] strs) {
        int n;
        if ((n = strs.length) == 0) return "";
        for (int i = 0; i < strs[0].length(); ++i) {
            for (int j = 1; j < n; ++j) {
                if (strs[j].length() <= i || strs[j].charAt(i) != strs[0].charAt(i)) {
                    return strs[0].substring(0, i);
                }
            }
        }
        return strs[0];
    }
}

C++

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        int n;
        if ((n = strs.size()) == 0) return "";
        for (int i = 0; i < strs[0].size(); ++i) {
            for (int j = 1; j < n; ++j) {
                if (strs[j].size() <= i || strs[j][i] != strs[0][i]) {
                    return strs[0].substr(0, i);
                }
            }
        }
        return strs[0];
    }
};

Go

func longestCommonPrefix(strs []string) string {
	if len(strs) == 0 {
		return ""
	}

	var b strings.Builder
	m, n := len(strs[0]), len(strs)

LOOP:
	for i := 0; i < m; i++ {
		for j := 1; j < n; j++ {
			if i >= len(strs[j]) || strs[0][i] != strs[j][i] {
				break LOOP
			}
		}
		b.WriteByte(strs[0][i])
	}

	return b.String()
}

...