[LeetCode]9.Palindrome Number

最新推荐文章于 2021-08-22 14:49:05 发布

bingbushangshu

最新推荐文章于 2021-08-22 14:49:05 发布

阅读量136

点赞数 1

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/bingbushangshu/article/details/91492430

版权

LeetCode 专栏收录该内容

7 篇文章

订阅专栏

本文探讨了判断数字是否为回文数的两种有效算法：逆序整数比较法和左右指针法。逆序整数比较法通过将数字逆序并与原数对比来判断；左右指针法则在不转换为字符串的情况下，从两端向中间对比数字的每一位。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true
Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:

Coud you solve it without converting the integer to a string?

解题方案

1 逆序整数

由于是数字，判断回文只需要求出倒过来的数字，判断两者是否相等，负数一定不是回文

bool isPalindrome(int num){

    int a=num;
    if (a<0)
        return false;
    //int yu=0;
    int temp=0;
    while(a){
        temp=10*temp+a%10;
        a=a/10;
    }
    if(num==temp){
        return true;
    }
    return false;
}

2 左右指针

左指向当前数字最高位，右指向当前数字最末尾

bool isPalindrome1(const int a) {
    if (a < 0)
        return false;
    int num = a;
    int n = 0;
    while (num != 0) {
        num /= 10;
        n++;
    }
    int left = 0;
    int right = 0;
    num = a;
    while (n > 1) {
        left = num / (pow(10, n - 1));
        right = num % 10;
        if (left != right) {
            return false;
        }
        num = (num - left * pow(10, n - 1) - right) / 10;
        n -= 2;
    }
    return true;
}