You are given two arrays (without duplicates)nums1
and nums2
where nums1
’s elements are subset of nums2
. Find all the next greater numbers for nums1
's elements in the corresponding places of nums2
.
The Next Greater Number of a number x in nums1
is the first greater number to its right in nums2
. If it does not exist, output -1 for this number.
Input:nums1 = [4,1,2], nums2 = [1,3,4,2]. Output: [-1,3,-1] Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Input:nums1 = [2,4], nums2 = [1,2,3,4]. Output: [3,-1] Explanation: For number 2 in the first array, the next greater number for it in the second array is 3. For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
- All elements in
nums1
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
implSolution{pubfnnext_greater_element(nums1:Vec<i32>,nums2:Vec<i32>) -> Vec<i32>{letmut ret = vec![-1; nums1.len()];for i in0..nums1.len(){for j in(0..nums2.len()).rev(){if nums2[j] > nums1[i]{ ret[i] = nums2[j];}elseif nums2[j] == nums1[i]{break;}}} ret }}
use std::collections::HashMap;implSolution{pubfnnext_greater_element(nums1:Vec<i32>,nums2:Vec<i32>) -> Vec<i32>{letmut stack = Vec::new();letmut num_greater = HashMap::new();for i in0..nums2.len(){while !stack.is_empty() && *stack.last().unwrap() < nums2[i]{ num_greater.insert(stack.pop().unwrap(), nums2[i]);} stack.push(nums2[i]);} nums1.iter().map(|x| *num_greater.get(x).unwrap_or(&-1)).collect()}}