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Determine whether an integer is a palindrome. Do this without extra space.
判断一个integer是否为回文。注意不要用额外的空间。
另外,注意考虑负数。
1 public class Solution { 2 public boolean isPalindrome(int x) { 3 if(x<0){ 4 return false; 5 } 6 7 int count=0; 8 int xx = x; 9 while(x!=0) { 10 x=x / 10; 11 count++; 12 } 13 // System.out.println(count); 14 15 int[] arr = new int[count]; 16 17 for(int i=0;i<count;i++) { 18 arr[i]=xx % 10; 19 20 xx /= 10; 21 // System.out.println(arr[i]); 22 } 23 24 int low = 0; 25 int high = count - 1; 26 while (low < high) { 27 if (arr[low] != arr[high]){ 28 return false; // Not a palindrome 29 } 30 low++; 31 high--; 32 } 33 34 return true; //is a palindrome 35 } 36 }
532 ms.
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原文地址:http://www.cnblogs.com/catcoding/p/4707341.html