leetcode1006

 func clumsy(N int) int {
     var ary []int
     for n := N; n >= ; n-- {
         ary = append(ary, n)
     }
     re := N %  //4个数字一组
     firstgroup := true
     //最后一组如果不够4个，就补充到4个值，*和/补充1，+和-补充0，这样结果不变
     if re ==  {
         //不用补充
     } else if re ==  {
         //补充3个
         ary = append(ary, ) //补充*1
         ary = append(ary, ) //补充/1
         ary = append(ary, ) //补充+0
     } else if re ==  {
         ary = append(ary, ) //补充/1
         ary = append(ary, ) //补充+0
     } else if re ==  {
         ary = append(ary, ) //补充+0
     }
     n := len(ary) //长度是4的倍数
     var sum int
     for a := ; a <= n-; a +=  {
         temp := ary[a] * ary[a+] / ary[a+]
         if firstgroup { //第一组的G1*G2/G3是正
             firstgroup = false
         } else {
             temp = temp * - //非第一组的G1*G2/G3是负
         }
         temp += ary[a+]
         sum += temp
     }
     return sum
 }

这道题的思路就是把运算映射到数组的index上，形成一种规律的运算方式，在循环中进行处理。