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

第一個解法，在 max 值之前都用stack解，max 之後用暴力解。

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
}