1523. Count Odd Numbers in an Interval Range

Easy

題目描述

Description

Given two non-negative integers low and high. Return the count of odd numbers between low and high (inclusive).

Example 1:

Input: low = 3, high = 7 Output: 3 Explanation: The odd numbers between 3 and 7 are [3,5,7].

Example 2:

Input: low = 8, high = 10 Output: 1 Explanation: The odd numbers between 8 and 10 are [9].

Constraints:

0 <= low <= high <= 10^9

解答

Solution

暴力解

/**
 * @param {number} low
 * @param {number} high
 * @return {number}
 */
var countOdds = function (low, high) {
  let result = 0;
  while (low <= high) {
    if (low % 2 === 1) {
      result++;
    }
    low++;
  }

  return result;
};

公式解

function countOdds(low: number, high: number): number {
  return Math.floor((high + 1) / 2) - Math.floor(low / 2);
}

解題思路

本來用 arr.lengh 作法結果看到記憶體爆炸，後來想通欸不對啊直接簡化用變數++就好就過了，但複雜度超糟就是ㄌ
hint中有提到一個點是可以從 (high - low + 1) 中判斷，改進之後變公式解甚至不用迴圈就可以直接算出。

心得

兩種解法分別放在 js 和 ts 檔案中，暴力解放 js，公式解放 ts，順便附上記憶體爆炸的紀念照