题解 | 两种排序方法

#include <iostream>
#include <string>
#include <vector>
using namespace std;

class Solution {
public:
    string checkOrder(vector<string>& strs) {
        bool isLex = true;  // 是否满足字典序
        bool isLen = true;  // 是否满足长度序
        
        // 检查相邻字符串
        for (int i = 1; i < strs.size(); i++) {
            // 检查字典序
            if (strs[i] <= strs[i-1]) {
                isLex = false;
            }
            // 检查长度序
            if (strs[i].length() <= strs[i-1].length()) {
                isLen = false;
            }
        }
        
        // 根据检查结果返回结果
        if (isLex && isLen) return "both";
        if (isLex) return "lexicographically";
        if (isLen) return "lengths";
        return "none";
    }
};

int main() {
    int n;
    cin >> n;
    
    vector<string> strs(n);
    for (int i = 0; i < n; i++) {
        cin >> strs[i];
    }
    
    Solution solution;
    cout << solution.checkOrder(strs) << endl;
    
    return 0;
}

import java.util.Scanner;

public class Main {
    static class Solution {
        public String checkOrder(String[] strs) {
            boolean isLex = true;  // 是否满足字典序
            boolean isLen = true;  // 是否满足长度序
            
            // 检查相邻字符串
            for (int i = 1; i < strs.length; i++) {
                // 检查字典序
                if (strs[i].compareTo(strs[i-1]) <= 0) {
                    isLex = false;
                }
                // 检查长度序
                if (strs[i].length() <= strs[i-1].length()) {
                    isLen = false;
                }
            }
            
            // 根据检查结果返回结果
            if (isLex && isLen) return "both";
            if (isLex) return "lexicographically";
            if (isLen) return "lengths";
            return "none";
        }
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        sc.nextLine();  // 消耗换行符
        
        String[] strs = new String[n];
        for (int i = 0; i < n; i++) {
            strs[i] = sc.nextLine();
        }
        
        Solution solution = new Solution();
        System.out.println(solution.checkOrder(strs));
        
        sc.close();
    }
}

class Solution:
    def check_order(self, strs: list) -> str:
        is_lex = True  # 是否满足字典序
        is_len = True  # 是否满足长度序
        
        # 检查相邻字符串
        for i in range(1, len(strs)):
            # 检查字典序
            if strs[i] <= strs[i-1]:
                is_lex = False
            # 检查长度序
            if len(strs[i]) <= len(strs[i-1]):
                is_len = False
        
        # 根据检查结果返回结果
        if is_lex and is_len:
            return "both"
        if is_lex:
            return "lexicographically"
        if is_len:
            return "lengths"
        return "none"

def main():
    n = int(input())
    strs = [input() for _ in range(n)]
    
    solution = Solution()
    print(solution.check_order(strs))

if __name__ == "__main__":
    main()