Palindrome Number
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回文数,当然有很简单的办法,那就是无比强大的 sprintf_s,但是题目限制不能使用额外空间,那么只好按部就班取数的每一位,然后再判断是否相等。
注意,负数无论如何也不能成为回文数,因为前面有负号。
按部就班版:
class Solution {public:bool isPalindrome(int x) {if (x == 0)return true;if (x < 0)return false;int safe = x;int count = 0;while (x) {x /= 10;count++;}int l = count - 1, r = 0;while (l >= r) {if (helper(safe, l) != helper(safe, r))return false;l--;r++;}return true;}int helper(int x, int i) {while (i > 0) {x /= 10;--i;}return x % 10;}};
sprintf_s版:
class Solution {public: bool isPalindrome(int x) { char data[13]; sprintf(data, "%d", x); int l = 0, r = strlen(data) - 1; while (l <= r) { if (data[l] != data[r]) return false; l++; r--; } return true; }};
http://oj.leetcode.com/problems/palindrome-number/
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