LeetCode Js-9. Palindrome Number

    AllenLi
    LeetCode Js-String

    Given an integer x, return true if x is a palindrome, and false otherwise.

    給予一個整數 x，如果 x 是一個回文的話，回傳 true，否則回傳 false。

    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.

    Solution:
    1. 回傳 x 的判斷式：
    a = x 的反轉後字串。
    b = x.的初始字串。
    return a === b
    ex. 0 只有一位數字，正向和反向讀都是 0。

    Code 1:

    var isPalindrome = function(x: number): boolean {
  if (x < 0) return false;
  if (x === 0) return true;
  return x.toString().split("").reverse().join("") === x.toString();
};

    FlowChart:
    Example 1

    Input: x = 121
console.log(x.toString().split('').reverse().join(''), x.toString())

return 121 === 121 //true

    Example 2

    Input: x = -121
console.log(x.toString().split('').reverse().join(''), x.toString())

return 121- === -121 //false

    Example 3

    Input: x = 10
console.log(x.toString().split('').reverse().join(''), x.toString())

return 10 === 01 //false

    reverse():

    var a = ['one', 'two', 'three'];
var reversed = a.reverse();

console.log(a);        // ['three', 'two', 'one']
console.log(reversed); // ['three', 'two', 'one']

    Code 2:

    var isPalindrome = function(x: number): boolean {
  if (x < 0) return false;
  if (x === 0) return true;
  let num: number = x,
    reverseNum: number = 0;
  while (num > 0) {
    reverseNum = reverseNum * 10 + (num % 10);
    num = Math.floor(num / 10);
  }
  return reverseNum === x;
};

    Code 3:

    var isPalindrome = function (x: number): boolean {
  if (x < 0) return false;
  if (x === 0) return true;
  return `${x}`.split("").reverse().join("") === `${x}`.toString();
};

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