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[LeetCode#9] Palindrome Number
The questions:
Determine whether an integer is a palindrome. Do this without extra space.
Analysis:
To determine whether an integer is a palindrome, we could try following ways:
1. reverse the num into num‘, test iff num == num‘. (This method may encounter complex overlfow handle)
2. use two pointer : front and end. Compare the digit at front and end, then move on !
Skill:
1. chop off the first digit, use ‘%‘
e: num = 65321, num‘ = num % 10000=65321 % 10000 = 5321, then the first digit was chopped off
2. chop off the last digit, use ‘/‘
e: num = 65321, num‘ = num / 10 = 6532, then the last digit was chopped off
Note: it‘s different from we want to get the first digit and last digit.
1. get the first digit, use "/"
e: num = 65321, digit = num / 10000=65321 / 10000 = 6
2. get the last digit, use "%"
e: num = 65321, num‘ = num % 10 = 2
My solution:
public class Solution { public boolean isPalindrome(int x) { if (x < 0) return false; int front_digit = 0; int end_digit = 0; int remain_number = x; int number_size = number_digit_size(x); int digit_left = number_size; while (digit_left > 1) { front_digit = remain_number / power_of_ten(digit_left - 1); end_digit = remain_number % 10; if (front_digit != end_digit) { return false; } remain_number = remain_number % power_of_ten(digit_left - 1);//chop the first digit of the number remain_number = remain_number / 10; //chop the last digit of the number digit_left = digit_left - 2; } return true; } static int power_of_ten(int count) { //return 10 ^ count int ret = 1; do { ret = ret * 10; count --; } while (count > 0); return ret; } static int number_digit_size(int number) { //get the size of digits in the number int size = 0; if (number == 0) { return 1; } while (number != 0) { number = number / 10; size ++; } return size; }}
[LeetCode#9] Palindrome Number