任務給定一個整數陣列，從陣列中找出兩個數之間的最大乘積，即 3 的倍數。
</script>
arr = [-9, -11, 4, 6, 9, 7];
arr2 = [-11, 4, 6, 7];
arr3 = [11, 3, 5]
function findProduct(arr) {
let maxProduct = 0;
for(let i = 0; i < arr.length - 1; i ) {
for(let j = i 1; j < arr.length; j ) {
let product = arr[i] * arr[j]
if(product % 3 !== 0) continue;
if(product > maxProduct) {
maxProduct = product
}
}
}
return maxProduct
}
console.log(findProduct(arr))
console.log(findProduct(arr2))
console.log(findProduct(arr3))
</script>
′′′′
uj5u.com熱心網友回復：
O(N)時間，O(1)額外空間演算法：
- 遍歷陣列，跟蹤兩個變數：
max_multiple_of_three和min_multiple_of_three - 遍歷陣列，跟蹤兩個變數：（
largest_number_in_array不應與 具有相同的索引max_multiple of three）和smallest_number_in_array（不應與 具有相同的索引min_multiple_of_three） - 答案將是
max_multiple_of_three * largest_number_in_array或者min_multiple_of_three * smallest_number_in_array
示例 1：
arr = [-9, -11, 4, 6, 9, 7]
max_multiple_of_three = 9
min_multiple_of_three = -9
largest_number_in_array = 7
smallest_number_in_array = -11
ans = max(-9*-11, 9*7) = 99
示例 2：
arr = [-11, 4, 6, 7]
max_multiple_of_three = 6
min_multiple_of_three = 6
largest_number_in_array = 7
smallest_number_in_array = -11
ans = max(6*-11, 6*7) = 42
轉載請註明出處，本文鏈接：https://www.uj5u.com/gongcheng/456437.html
上一篇：在foreach回圈中爆炸（）