LeetCode9-Palindrome Number
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【题目】
Determine whether an integer is a palindrome. Do this without extra space.
判断一个整型数字是否为回文,要求空间复杂度为O(1)。
【思路】
1、首先判断该数字的位数,判断方法例如:999<=x<=9999,则说明x为4位数;
2、i从高位到低位,j从低位到高位,依次判断第i位和第j位是否相等,直至i<j;
【Java代码】
public class Solution_9_is_palindrome_num {public boolean isPalindrome(int x){boolean result = false;if(x >= 0){int digit = 1;int flag = 0;int[] judge_digit = {-1,9,99,999,9999,99999,999999,9999999,99999999,999999999};for(int i = 9 ;; i--){if(x > judge_digit[i]){digit = i+1; break;}}for(int i = digit,j=1 ;i>=j; i--,j++){if(getdigit(x,i) != getdigit(x,j)){flag = 1;break;}}if(flag == 0)result = true;}return result;}public int getdigit(int x, int index){index = index - 1;x = x/(int)Math.pow(10, index);return x%10;}}
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