algorithm - 在数组中获取与索引相反的元素的最佳方法是什么？

Question

说我有数组

{0, 1, 2, 3, 4}

• 索引 0 和 4 是相反的
• 索引 1 和 3 是相反的
• 索引 2 的对面未定义

或者

 {0, 1, 2, 3, 4, 5}

• 索引 0 和 5 是相反的
• 索引 1 和 4 是相反的
• 索引 2 和 3 是相反的

我记得看到一个非常聪明的方法来做到这一点。就像是

i%array.length 

Answer 1

试试这个：

oppIndex = array.length - firstIndex - 1;

Answer 2

array = {1, 2, 3, 4, 5}
idx = 0 /* 0 is the first array position with value of 1 */

IF array.length - idx - 1 > idx
   RETURN array[ array.length - idx - 1 ] /* returns array[4] == 5 */
ELSE
   RETURN undefined

array = {1, 2, 3, 4, 5}
idx = 2 /* 2 is the third array position with value of 3 */

IF array.length - idx - 1 > idx /* 5 - 2 - 1 == 2 which is NOT greater than 2 */
   RETURN array[ array.length - idx - 1 ]
ELSE
   RETURN undefined

Answer 3

按照您的第一个示例逻辑...

       if (array.length%2==0){
            for (int i =0; i<array.length/2; i++){
                   syso("the opposite of "+ array[i] + " is " + array[array.length -1-i]
             }
       }
       else {
           for (int i =0; i<floor(array.length/2); i++){
                   syso("the opposite of "+ array[i] + " is " + array[array.length -1-i]
             }
             i = i+1
              syso("the opposite of "+ array[i] +" is undefined"
       }