LeetCode开心刷题第七天——13RomanToInteger14 Longest Common Prefix

13. Roman to Integer

Easy

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Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9. 
• X can be placed before L (50) and C (100) to make 40 and 90. 
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
this problem is easy level.so I can finish it easily,but the switch case method is too redundant,So i find a new way,the most remarkable thing
is the dict array.The order of the array is from large to small .This is the rule of Roman Number.

class Solution {
public:
    int romanToInt(string s) {
        string dict[] = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
        int num[] = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
        int i=0, index=0, ret=0;
        while(i<s.size() && index<13) {
            string target = dict[index];
            string cur = s.substr(i,target.size());
            if(cur==target) {
                ret += num[index];
                i += target.size();
            }
            else {
                index++;
            }
        }
        return ret;
    }
};

14. Longest Common Prefix

Easy

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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:

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

Example 2:

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.

Analyse：This is also a easy question.one remarkable point is that char* is a really useful tool compare to string.Because String when we assign value to string[i]=0 \0.The length of string won't be changed .But in the situation of char* when we assign char*[i]=0.The length will stop in the point of i.This is a useful property when we code.

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h>
#include<iomanip>
#include<vector>
#include<list>
#include<queue>
#include<algorithm>
#include<map>
using namespace std;

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        if(strs.size()==0) return "";
        int size1=strs[0].size(),j=0;
        char* str=(char*)malloc(sizeof(char)*(size1+1));
        for(int i=0;i<size1;i++)
        {
            str[i]=strs[0][i];
        }
        str[size1]=0;
        for(int i=1;i<strs.size();i++)
        {
            j=0;
            while(str[j]&&strs[i][j]&&str[j]==strs[i][j]) j++;
            str[j]=0;
        }
        return str;
    }
};
int main()
{
    Solution s;
    string temp,res;
    char ch1;
    vector<string> str;
    while((ch1=cin.get())!='\n' )
    {
        cin.putback(ch1);
        cin>>temp;
        str.push_back(temp);
    }
    res=s.longestCommonPrefix(str);
    cout<<res<<endl;
    return 0;
}

转载于:https://www.cnblogs.com/Marigolci/p/11064246.html