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9_Palindrome Number
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9.Palindrome Number
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: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: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?
版本一: 思想是把数字拆成一位一位的数组, 首位比较, 好像转换成字符串了, 违反Follow up了
// 偶数位时回文数的退出是关键,
class Solution {
public:
bool isPalindrome(int x) {
bool ret = false;
vector<int> position;
int begin = 0, end = 0;
if (x < 0 ) {
return ret;
}
else if (0 == x) {
return true;
}
else {
while (0 != x) {
position.push_back(x % 10);
x /= 10;
}
end = position.size() - 1;
while (begin != end) {
if (position[begin] != position[end]) {
return ret;
}
// 偶数个位数时终止循环, 否则判断之后数组访问越界, 位置不能换
if ((begin + 1) == end) {
break;
}
begin++;
end--;
}
ret = true;
return ret;
}
}
};
版本二: 借助7_Reverse Integer翻转数字的思想, 比较原数字和翻转之后的数字是否相同进行判断
// 借助[7_Reverse Integer](https://www.cnblogs.com/hesper/p/10397725.html)翻转数字的思想
// 比较原数字和翻转之后的数字是否相同进行判断
class Solution {
public:
bool isPalindrome(int x) {
int y = x;
//int tmp = 0; // int 类型x翻转 后数字范围可能大于int最大值
long tmp = 0;
vector<int> position;
int begin = 0, end = 0;
if (x < 0 ) {
return false;
}
else {
while (0 != x) {
tmp = tmp*10 + x%10;
x /= 10;
}
if (tmp == y) {
return true;
}
else {
return false;
}
}
}
};
LeetCode精简版
class Solution {
public:
bool isPalindrome(int x) {
if (x < 0)
return false;
long long result = 0;
int temp = x;
while(temp) {
result *= 10;
result += temp % 10;
temp /= 10;
}
if ((int)result == x)
return true;
return false;
}
};
