9. Palindrome Number.md

9. Palindrome Number

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true

Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

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Solution

• mine
  • Java

    • String Runtime: 11 ms, faster than 23.00%, Memory Usage: 42.8 MB, less than 5.01% of Java online submissions

      // O(N)time O(N)space  
      // N is x's length
      public boolean isPalindrome(int x) {
          if(x < 0  || (x % 10 == 0 && x != 0)){
              return false;
          }
          String s = String.valueOf(x);
          char[] array = s.toCharArray();
          int i = 0;
          while(i <=  array.length - 1 - i){
              if(array[i] != array[array.length - 1 - i]){
                  return false;
              }
              i++;
          }
          return true;
      }
      

    • Math Runtime: 6 ms, faster than 100.00%, Memory Usage: 38.8 MB, less than 64.50% of Java online submissions

      // O(N)time O(1)space  
      // N is x's length
      public boolean isPalindrome(int x) {
          if(x < 0 || (x != 0 && x % 10 == 0)){
              return false;
          }
          int t = x;
          int res = 0;
          while(t > 0){
              res = res * 10 + t % 10;
              t = t / 10;
          }
          return res == x;
      }
      

    • LinkedList Runtime: 17 ms, faster than 9.62%, Memory Usage: 45 MB, less than 5.01% of Java online submissions

      // O(N)time O(N)space  
      // N is x's length
      public boolean isPalindrome(int x) {
          if (x < 0) {
              return false;
          } else if (x < 10) {
              return true;
          } else if (x % 10 == 0) {
              return false;
          }
          LinkedList<Integer> list = new LinkedList<>();
          while (x > 0) {
              list.add(x % 10);
              x /= 10;
          }
          while (list.size() > 1) {
              if (!list.removeFirst().equals(list.removeLast())) {
                  return false;
              }
          }
          return true;
      }
      

    • Math Runtime: 9 ms, faster than 33.82%, Memory Usage: 41.9 MB, less than 6.33% of Java online submissions

      // O(N)time O(1)space  
      // N is x's length
      public boolean isPalindrome(int x) {
          if (x < 0) {
              return false;
          } else if (x < 10) {
              return true;
          } else if (x % 10 == 0) {
              return false;
          }
          int max = 1;
          while (x / max > 9) {
              max *= 10;
          }
          while (max >= 10 && x / max == x % 10) {
              x = x % max;
              x /= 10;
              max /= 100;
          }
          return max < 10;
      }
      

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• the most votes
  • Math Runtime: 8 ms, faster than 41.65%, Memory Usage: 38.6 MB, less than 85.60% of Java online submissions

    //O(N)time O(1)space
    // N is x's length
    public boolean isPalindrome(int x) {
        if (x < 0 || (x != 0 && x % 10 == 0)) return false;
        int rev = 0;
        while (x > rev) {
            rev = rev * 10 + x % 10;
            x = x / 10;
        }
        return (x == rev || x == rev / 10);
    }
    

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• leetcode solution

  • Revert half of the number

    Runtime: 6 ms, faster than 100.00%, Memory Usage: 38.8 MB, less than 64.50% of Java online submissions

    //O(N)time O(1)space
    // N is x's length
    public boolean isPalindrome(int x) {
        if (x < 0 || (x % 10 == 0 && x != 0)) {
            return false;
        }
    
        int revertedNumber = 0;
        while (x > revertedNumber) {
            revertedNumber = revertedNumber * 10 + x % 10;
            x /= 10;
        }
        return x == revertedNumber || x == revertedNumber / 10;
    }
    

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