# 503. Next Greater Element II

Medium
Given a circular integer array `nums` (i.e., the next element of `nums[nums.length - 1]` is `nums[0]`), return the next greater number for every element in `nums`.
The next greater number of a number `x` is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, return `-1` for this number.
Example 1:
Input: nums = [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number.
The second 1's next greater number needs to search circularly, which is also 2.
Example 2:
Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]
Constraints:
• `1 <= nums.length <= 10^4`
• `-10^9 <= nums[i] <= 10^9`

### 解題

func nextGreaterElements(nums []int) []int {
stack := make([]int, 0)
stack = append(stack, 0)
ans := make([]int, len(nums))
for i:=1; i<len(nums); i++ {
for len(stack) != 0 && nums[stack[len(stack)-1]] < nums[i] {
ans[stack[len(stack)-1]] = nums[i]
stack = stack[:len(stack)-1]
}
stack = append(stack, i)
}
ans[stack[0]] = -1
max := stack[0]
stack = stack[1:]
// 後面暴力解
for i:=0; i<len(stack); i++ {
for j:=0; j<=max; j++ {
if nums[j] > nums[stack[i]] {
ans[stack[i]] = nums[j]
break
}
if j==max{
ans[stack[i]] = -1
}
}
}
return ans
}
Last modified 11mo ago