Leetcode No. Palindrome Number回文数(c++实现)

1. 题目

https://leetcode.com/problems/palindrome-number/

2. 分析

这一题可以将数字直接反转,如果反转后数字和原数字相同,则一定是回文数。当然对于负数可以直接返回false
具体代码如下:

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        int temp = x;
        int64_t reverse = 0;
        while (temp != 0) {
            reverse = reverse * 10 + temp % 10;
            temp /= 10;
        }
        return reverse == x ? true : false;
    }
};

但是这种算法不够优化,可以只反转后半部分,之后与前一半进行对比即可。
具体代码如下:

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0 || (x != 0 && x % 10 == 0)) {
            return false;
        }
        int reverseHalf = 0;//将x的后一半进行反转,并存储在reverseHalf中
        while (reverseHalf < x) {//x只留下前一半(如果x为奇数,中间那一位给后一半)
            reverseHalf = reverseHalf * 10 + x % 10;
            x /= 10;
        }
        return reverseHalf == x || reverseHalf / 10 == x;
    }
};

参考:https://leetcode.com/problems/palindrome-number/discuss/5165/An-easy-c%2B%2B-8-lines-code-(only-reversing-till-half-and-then-compare)

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