寫(xiě)在前面:

程序=數(shù)據(jù)結(jié)構(gòu)+算法，數(shù)據(jù)結(jié)構(gòu)與算法的重要性就不多說(shuō)了。幾乎沒(méi)有一個(gè)一線互聯(lián)網(wǎng)公司招聘任何類(lèi)別的技術(shù)人員是不考算法的，程序猿們都懂的，現(xiàn)在最權(quán)威流行的刷題平臺(tái)就是 LeetCode。

LeetCode原題鏈接

string - C++ Reference

C++中int與string的相互轉(zhuǎn)換

C++ Map常見(jiàn)用法說(shuō)明

Question(Easy):

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

Example:

Input: 121
Output: true

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.

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?

Solution

以下代碼皆是本人用 C++寫(xiě)的，覺(jué)得還不錯(cuò)的話別忘了點(diǎn)個(gè)贊哦。各位同學(xué)們?nèi)绻衅渌咝Ы忸}思路還請(qǐng)不吝賜教，多多學(xué)習(xí)。

A1、反轉(zhuǎn)數(shù)字判斷回文數(shù)

回文數(shù)：1.負(fù)數(shù)排除 ；2.先反轉(zhuǎn)數(shù)字，再比較是否與原數(shù)相等

參考LeetCode刷算法題 - 7. Reverse Integer

算法時(shí)間復(fù)雜度 O(logn)，Runtime: 136 ms，代碼如下

static int x=[](){
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}();

class Solution {
public:
    bool isPalindrome(int x) {
        if (x<0) {
            return false;
        }
        long res = 0;
        long temp = x;
        while (temp) {
            res = res*10 + temp%10;
            temp /= 10;
        }
        return res == x;
    }
};

引用相關(guān)

LeetCode 官網(wǎng)