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9.palindrome-number

Statement

Metadata
  • Link: 回文数
  • Difficulty: Easy
  • Tag: 数学

给你一个整数 x ，如果 x 是一个回文整数，返回 true ；否则，返回 false 。

回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。例如，121 是回文，而 123 不是。

 

示例 1：

输入：x = 121
输出：true

示例 2：

输入：x = -121
输出：false
解释：从左向右读, 为 -121 。 从右向左读, 为 121- 。因此它不是一个回文数。

示例 3：

输入：x = 10
输出：false
解释：从右向左读, 为 01 。因此它不是一个回文数。

示例 4：

输入：x = -101
输出：false

 

提示：

  • -231 <= x <= 231 - 1

 

进阶：你能不将整数转为字符串来解决这个问题吗？

Metadata

Given an integer x, return true if x is palindrome integer.

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

  • For example, 121 is a palindrome while 123 is not.

 

Example 1:

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

Example 2:

Input: x = -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: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

 

Constraints:

  • -231 <= x <= 231 - 1

 

Follow up: Could you solve it without converting the integer to a string?

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define lowbit(x) ((x) & (-(x)))
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) {
            return false;
        }

        string s = to_string(x);
        int len = s.size();
        for (int i = 0, j = len - 1; i < j; ++i, --j) {
            if (s[i] != s[j]) {
                return false;
            }
        }

        return true;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif
最后更新: October 11, 2023
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