算法分析與設計——LeetCode Problem.12 Integer to Roman

題目鏈接

問題描述

Given an integer, convert it to a roman numeral.

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

解題思路

將1000,900,500,400,100,90,50,40,10,9,5,4,1與對應的羅馬數字用map存儲，然後根據輸入的數字輸出相應的羅馬數字即可

代碼如下：

class Solution {
	map<int, string> iToR;
public:
	string findT(int num) {
		string s = "";
		string tempstr;
		for (int i = 0; i < num; i++) {
			tempstr = iToR.find(1000)->second;
			s = s + tempstr;
		}
		return s;
	}

	string findH(int num) {
		string s = "";
		string tempstr;
		if (num == 0) {
			return s;
		}
		if (num < 4) {
			for (int i = 0; i < num; i++) {
				tempstr = iToR.find(100)->second;
				s = s + tempstr;
			}
			return s;
		}
		if (num == 4) {
			tempstr = iToR.find(400)->second;
			s = s + tempstr;
			return s;
		}
		if (num < 9) {
			int temp = num - 5;
			tempstr = iToR.find(500)->second;
			s = s + tempstr;
			for (int i = 0; i < temp; i++) {
				tempstr = iToR.find(100)->second;
				s = s + tempstr;
			}
			return s;
		}
		if (num == 9) {
			tempstr = iToR.find(900)->second;
			s = s + tempstr;
			return s;
		}
	}

	string findTen(int num) {
		string s = "";
		string tempstr;
		if (num == 0) {
			return s;
		}
		if (num < 4) {
			for (int i = 0; i < num; i++) {
				tempstr = iToR.find(10)->second;
				s = s + tempstr;
			}
			return s;
		}
		if (num == 4) {
			tempstr = iToR.find(40)->second;
			s = s + tempstr;
			return s;
		}
		if (num < 9) {
			int temp = num - 5;
			tempstr = iToR.find(50)->second;
			s = s + tempstr;
			for (int i = 0; i < temp; i++) {
				tempstr = iToR.find(10)->second;
				s = s + tempstr;
			}
			return s;
		}
		if (num == 9) {
			tempstr = iToR.find(90)->second;
			s = s + tempstr;
			return s;
		}
	}

	string findOne(int num) {
		string s = "";
		string tempstr;
		if (num == 0) {
			return s;
		}
		if (num < 4) {
			for (int i = 0; i < num; i++) {
				tempstr = iToR.find(1)->second;
				s = s + tempstr;
			}
			return s;
		}
		if (num == 4) {
			tempstr = iToR.find(4)->second;
			s = s + tempstr;
			return s;
		}
		if (num < 9) {
			int temp = num - 5;
			tempstr = iToR.find(5)->second;
			s = s + tempstr;
			for (int i = 0; i < temp; i++) {
				tempstr = iToR.find(1)->second;
				s = s + tempstr;
			}
			return s;
		}
		if (num == 9) {
			tempstr = iToR.find(9)->second;
			s = s + tempstr;
			return s;
		}
	}

	string intToRoman(int num) {
		//map<int, string> iToR;
		iToR.insert(pair<int, string>(1, "I"));
		iToR.insert(pair<int, string>(4, "IV"));
		iToR.insert(pair<int, string>(5, "V"));
		iToR.insert(pair<int, string>(9, "IX"));
		iToR.insert(pair<int, string>(10, "X"));
		iToR.insert(pair<int, string>(40, "XL"));
		iToR.insert(pair<int, string>(50, "L"));
		iToR.insert(pair<int, string>(90, "XC"));
		iToR.insert(pair<int, string>(100, "C"));
		iToR.insert(pair<int, string>(400, "CD"));
		iToR.insert(pair<int, string>(500, "D"));
		iToR.insert(pair<int, string>(900, "CM"));
		iToR.insert(pair<int, string>(1000, "M"));

		int thous = num / 1000;
		int huns = (num / 100) % 10;
		int tens = (num / 10) % 10;
		int ones = num % 10;
		string s;
		s = findT(thous) + findH(huns) + findTen(tens) + findOne(ones);
		return s;
	}
};