题目说明:
Determine whether an integer is a palindrome. Do this without extra space.
Some hints:
Could negative integers be palindromes? (ie, -1)
If you are thinking of converting the integer to string, note the restriction of using extra space.
You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?
There is a more generic way of solving this problem.
程序代码:
#include <gtest/gtest.h> using namespace std; bool isPalindrome2(int x) { bool bResult = false; if (x < 0) { return false; } int tempData[20] = {0}; int tempIdx = 0; int tempX = x; long long rValue = 0; while (tempX) { tempData[tempIdx++] = tempX % 10; tempX /= 10; } for (int i=0; i<tempIdx;++i) { rValue = rValue*10 +tempData[i]; } return (x == rValue); } bool isPalindrome3(int x) { if (x < 0) return false; long long nValue = 0; int temp = x; while (temp) { nValue = nValue*10 + temp % 10; temp /= 10; } return (x == nValue); } bool isPalindrome(int x) { if (x < 0) return false; int dev = 1; while (x / dev >= 10) { dev *= 10; } while (x != 0) { int l = x / dev; int r = x % 10; if (l != r) return false; x = (x % dev) / 10; dev /= 100; } return true; } TEST(Pratices, tIsPalindrome) { // 123 false // 121 true // -111 false // 0 true // 2147483647 false ASSERT_FALSE(isPalindrome(123)); ASSERT_TRUE(isPalindrome(121)); ASSERT_FALSE(isPalindrome(-111)); ASSERT_TRUE(isPalindrome(0)); ASSERT_FALSE(isPalindrome(2147483647)); }
本文提供了一种不使用额外空间的方法来判断整数是否为回文数,包括处理负数和溢出情况,并通过测试用例验证了算法的有效性。
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