Roman to Integer(罗马数转为整数)leetcode13

Roman numerals are represented by seven different symbols: IVXLCD 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: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

code1:

    public int romanToInt(String s) {
        /*int[] nums={1000,900,500,400,100,90,50,40,10,9,5,4,1};
        String[] romans={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
        int count=0;
        int result=0;
        while(count<s.length()){
            String res1="";
            if(count<s.length()-1)
                res1=s.substring(count,count+2);
            String res2=s.substring(count,count+1);
            for(int i=0;i<romans.length;i++){
                if(romans[i].equals(res1)){
                    count+=2;
                    result+=nums[i];
                    break;
                }else if(romans[i].equals(res2)){
                    count++;
                    result+=nums[i];
                    break;
                }
            }
            
        }
        return result;
}

code2:

 public int romanToInt(String s) {
        int nums[]=new int[s.length()];
    for(int i=0;i<s.length();i++){
        switch (s.charAt(i)){
            case 'M':
                nums[i]=1000;
                break;
            case 'D':
                nums[i]=500;
                break;
            case 'C':
                nums[i]=100;
                break;
            case 'L':
                nums[i]=50;
                break;
            case 'X' :
                nums[i]=10;
                break;
            case 'V':
                nums[i]=5;
                break;
            case 'I':
                nums[i]=1;
                break;
        }
    }
    /*int sum=0;//从前往后遍历或者下面那种从后往前遍历
    for(int i=0;i<nums.length-1;i++){
        if(nums[i]<nums[i+1])
            sum-=nums[i];
        else
            sum+=nums[i];
    }
    return sum+nums[nums.length-1];*/
        int sum=nums[nums.length-1];
        for(int i=nums.length-2;i>=0;i--){
            if(nums[i]<nums[i+1])
                sum-=nums[i];
            else
                sum+=nums[i];
        }
        return sum;
    }

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转载自blog.csdn.net/m0_37402140/article/details/80387102