LeetCode-9. Palindromes (day2)

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I. Preface:

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Ii. Title Description:

Difficulty: ⭐ ⭐

Given an integer x, return true if x is a palindrome integer; Otherwise, return false.

Palindromes are integers that read in positive (left to right) and backward (right to left) order.

For example,121It’s a palindrome, and123It isn’t.

Example 1:

Input: x = 121 Output: trueCopy the code

Example 2:

Input: x = -121 Output: false Description: Read from left to right. The value is -121. Read from right to left, 121-. So it's not a palindrome number.Copy the code

Title: LeetCode website

Iii. Analysis of Ideas:

Idea 1: Turn to string method

The best thing to do, as you can imagine, is to turn the integer into a string, then split the string into an array, and then loop through half the length of the array to see if the elements are equal.

Idea 2: Number flip method

When we look at palindromes intuitively, it’s like folding numbers in half to see if they match. So the operation of this solution is to take the last part of the number and flip it.

One point that needs to be noted here is that the number of palindromes can be even and odd, so please see the following:

When its length is even, it has to be equal when folded in half;
When it has an odd length, then when it’s folded in half, the longer half needs to get rid of the last digit, because it’s flipped, so get rid of the last digit, which is the first digit in the second half.

Iv. Algorithm implementation:

Turn string method _AC code

The specific implementation code is as follows:

Class Solution {public Boolean isPalindrome(int x) {public Boolean isPalindrome(int x) {String STR = string.valueof (x); For (int I = 0; i < str.length() / 2; If (str.charat (I)!) {// If (str.charat (I)! = str.charAt(str.length() - i - 1)) { return false; } } return true; }}Copy the code

Number flipping method _AC code

The specific implementation code is as follows:

Class Solution {public Boolean isPalindrome(int x) {public Boolean isPalindrome(int x) {class Solution {public Boolean isPalindrome(int x) { if (x < 0 || (x % 10 == 0 && x ! = 0)) { return false; } // revertNum = 0; While (x > revertNum) {revertNum = revertNum * 10 + x % 10; x /= 10; } return x == revertNum || x == revertNum / 10; }}Copy the code

V. Summary:

The results of leetcode submission are as follows:

To sum up, the two methods have their own angles. Although the first way is often thought of, it is best to solve the problem according to one way. However, we still need to make a comparison. The second idea is to directly make use of the characteristics of palindromes. We only need to take a half-flip comparison and then start from a mathematical perspective to intercept numbers, which obviously improves the operation efficiency of the algorithm.

. .

Furthermore, there are thousands of ways to solve the problem. If you have any better ideas or ideas, please let me know in the comment section. We can learn from each other and grow faster.

Well, that’s all for this episode and I’ll see you next time.

Six, the previous recommendation:

Leetcode -1. Sum of two numbers
Leetcode – 9. Palindrome

. .

Vii. Finally:

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