LeetCode #11 简单题(给定了范围,直接硬编码就好了。。)

              题目:整数转罗马数字

              题解:范围限定在0-3999,硬编码就好- -懒得想逻辑了

              class Solution {
              public:
                  string intToRoman(int num) {
                      int a = num / 1000;
                      int b = (num / 100) % 10;
                      int c = (num / 10) % 10;
                      int d = num % 10;
                      string ans;
                      for (int i = 0; i < a; ++i)ans = ans + "M";
                      if (b == 9)ans = ans + "CM";
                      else if (b == 8)ans = ans + "DCCC";
                      else if (b == 7)ans = ans + "DCC";
                      else if (b == 6)ans = ans + "DC";
                      else if (b == 5)ans = ans + "D";
                      else if (b == 4)ans = ans + "CD";
                      else if (b == 3)ans = ans + "CCC";
                      else if (b == 2)ans = ans + "CC";
                      else if (b == 1)ans = ans + "C";
              
                      if (c == 9)ans = ans + "XC";
                      else if (c == 8)ans = ans + "LXXX";
                      else if (c == 7)ans = ans + "LXX";
                      else if (c == 6)ans = ans + "LX";
                      else if (c == 5)ans = ans + "L";
                      else if (c == 4)ans = ans + "XL";
                      else if (c == 3)ans = ans + "XXX";
                      else if (c == 2)ans = ans + "XX";
                      else if (c == 1)ans = ans + "X";
              
                      if (d == 9)ans = ans + "IX";
                      else if (d == 8)ans = ans + "VIII";
                      else if (d == 7)ans = ans + "VII";
                      else if (d == 6)ans = ans + "VI";
                      else if (d == 5)ans = ans + "V";
                      else if (d == 4)ans = ans + "IV";
                      else if (d == 3)ans = ans + "III";
                      else if (d == 2)ans = ans + "II";
                      else if (d == 1)ans = ans + "I";
              
                      return ans;
                  }
              };
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