leetcode-14-Longest-Common-Prefix

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描述

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:

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

Example 2:

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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.

分析

题意是有一堆字符串，求这些字符串中的最长公共前缀，可以将第一个字符串作为操作对象，比较第 n 个字符，只要出现一个字符串的第 n 个字符不等于第一个字符串的第 n 个字符，就返回第一个字符串的 0~n-1 子串。

解决方案1（Python）

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class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        if not strs:
            return ""
        
        length, count = len(strs[0]), len(strs)
        
        for i in range(length):
            nowCh = strs[0][i]
            if any(i == len(strs[j]) or strs[j][i] != nowCh for j in range(1, count)):
                return strs[0][:i]
        return strs[0]

解决方案1(C++)

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

解决方案2(Rust)

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pub struct Solution {}

use std::str::Chars;
impl Solution {
    pub fn longest_common_prefix(strs: Vec<String>) -> String {
        let mut prefix = String::new();
        let mut str_iters: Vec<Chars> = strs.iter().map(|s| {s.chars()}).collect();
        let mut current_char: Option<char> = None;
        if strs.len() < 1 {
            return prefix
        }

        loop {
            current_char.take().map(|ch| prefix.push(ch));
            for iter in str_iters.iter_mut() {
                let mut ch = iter.next();
                if ch.is_none() {
                    return prefix
                }
                match current_char {
                    None => current_char = ch.take(),
                    Some(curr) => {
                        if curr != ch.unwrap() {
                            return prefix
                        }
                    }
                }
            }
        }
    }
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_leetcode_14() {
        assert_eq!(Solution::longest_common_prefix(
            vec![
                "flower".to_string(),
                "flow".to_string(),
                "flight".to_string()
            ]), "fl");
        assert_eq!(Solution::longest_common_prefix(
            vec![
                "dog".to_string(),
               "racecar".to_string(),
                "car".to_string()
            ]), "");
        assert_eq!(Solution::longest_common_prefix(vec![]), "");
    }
}

解决方案3（Golang）

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func longestCommonPrefix(strs []string) string {
    strsCount := len(strs)
    if strsCount == 0 {
        return ""
    }
    
    var result = len(strs[0])
    
    for nowStrIndex := 0; nowStrIndex < strsCount-1; nowStrIndex++ {
        nowEnd := 0
        nowStrLen := len(strs[nowStrIndex+1])
        if nowStrLen > result {
            nowStrLen = result
        }
        for i := 0; i < nowStrLen; i++ {
            if strs[nowStrIndex][i] == strs[nowStrIndex+1][i] {
                nowEnd++
            } else {
                break
            }
        }
        if result >= nowEnd {
            result = nowEnd
        }
    }
    return strs[0][0: result]
}

题目来源