leetcode1006

func clumsy(N int) int {
    var ary []int
    for n := N; n >= 1; n-- {
        ary = append(ary, n)
    }
    re := N % 4 //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 == 0 {
        //不用补充
    } else if re == 1 {
        //补充3个
        ary = append(ary, 1) //补充*1
        ary = append(ary, 1) //补充/1
        ary = append(ary, 0) //补充+0
    } else if re == 2 {
        ary = append(ary, 1) //补充/1
        ary = append(ary, 0) //补充+0
    } else if re == 3 {
        ary = append(ary, 0) //补充+0
    }
    n := len(ary) //长度是4的倍数
    var sum int
    for a := 0; a <= n-4; a += 4 {
        temp := ary[a] * ary[a+1] / ary[a+2]
        if firstgroup { //第一组的G1*G2/G3是正
            firstgroup = false
        } else {
            temp = temp * -1 //非第一组的G1*G2/G3是负
        }
        temp += ary[a+3]
        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。