503. 下一个更大元素 II

需要计算需要遍历的索引位置是
idx :=(j+i)%len(nums)

package main
import "fmt"
func nextGreaterElements(nums []int) []int {
     res :=make([]int,len(nums))
     flag :=false
     for i:=0;i<len(nums);i++{
         flag = false
         for j:=1;j<len(nums);j++{
              idx :=(j+i)%len(nums)
             if nums[i]<nums[idx] {
                  res[i] = nums[idx]
                  flag = true
                break
             }
         }
         if !flag {
             res[i] = -1
         }
     }
     return res
}
func main() {
    fmt.Println(nextGreaterElements([]int{1,2,1}))
    fmt.Println(nextGreaterElements([]int{5,4,3,2,1}))
}