LeetCode12~14 Integer to Roman/Roman to Integer/Longest Common Prefix

一:Integer to Roman

题目:

Given an integer, convert it to a roman numeral.

Input is guaranteed to be within the range from 1 to 3999.

链接:https://leetcode.com/problems/integer-to-roman/

分析:此题关键是确定罗马数字如何表示,比如4并不是IIII而是IV,9并不是VIIII而是IX, 通过不断的取出最高位和余下的得到结果

代码:

class Solution {
public:
    // 关键在于4 9 400 等表示方式的不同
    string intToRoman(int num) {
        string roman[13] ={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};  // 确定相互之间的对应关系
        int arabic[13] = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
        string result;
        for(int i = 0; i < 13; i++){
            int fac = num/arabic[i];   // 取出最高位,不是按照十进制/10
            for(int j = 0; j < fac; j++){
                result += roman[i];
            }
            num = num % arabic[i];   // 余下的
        }
        return result;

    }

};

二:

Roman to Integer

题目:

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

链接:https://leetcode.com/problems/roman-to-integer/

分析:这个更简单,只要分析当前字符串中字符位与后面一位对应大小关系判断是否为4,9等这样的特殊情况进行判断就可以了。

class Solution {
public:
    Solution(){
        string roman ={"MDCLXVI"};
        int arabic[7] = {1000,500,100,50,10,5,1};
        for(int i = 0; i < 7; i++){
            hmap.insert(pair<char,int>(roman[i], arabic[i]));    // 确定两者对应关系
        }
    }
    int romanToInt(string s) {
       int result = 0;
       int i = 0;
       if(s.size() == 0) return 0;
       if(s.size() == 1) return hmap[s[0]];
       while(i < s.size()){
            int x1 = hmap[s[i]];
            int x2 = hmap[s[i+1]];
            if(x1 >= x2){                   // 大于 正常
                result += x1;
                i = i+1;
            }
            else {                       // 小于 不正常说明是4 9 400 900 等这些数字
                result += x2-x1;
                i = i+2;
            }
        }
        return result;
    }
private:
    map<char, int> hmap;
};

三:Longest Common Prefix

题目:

Write a function to find the longest common prefix string amongst an array of strings.

链接:https://leetcode.com/problems/longest-common-prefix/

分析:此题就是求最长公共前缀,这个可以采用hash_set, 用第一个字符串的所有前缀建立一个set,然后对于后面每个字符串,看他的前缀是否在hash_set中,不在则判断是否比之前的preCommon小,小则取最短的

代码:

class Solution {
public:
    string longestCommonPrefix(vector<string> &strs) {
        if(strs.size() <= 0) return "";
        set<string> hset;
        string str0 = strs[0];
        int preCommon =  str0.size();
        for(int i = 0; i < str0.size(); i++){  // 第一个字符串的前缀建立一个hash_set
            hset.insert(str0.substr(0,i+1));
        }
        for(int i = 1; i < strs.size(); i++){
            string str = strs[i];
             preCommon = min(preCommon, (int)str.size());   // 公共前缀肯定是字符串中最短的
             if(preCommon == 0) break;
            for(int j = 0; j < str.size(); j++){
                if(!hset.count(str.substr(0,j+1))){         // 判断是否为前面字符串的前缀 O(1)时间搞定
                   preCommon = min(preCommon, j);      // hash_set中不存在了 则为最短
                   break;
                }
            }
        }
        return str0.substr(0, preCommon);
    }
};
时间: 2024-11-08 19:27:36

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