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.
Example 1:
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.
Example 2:
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.
Note:
- All elements in
nums1
andnums2
are unique. - The length of both
nums1
andnums2
would not exceed 1000.
在nums2对应的位置查找nums1的所有元素的下一个更大的数字。
C++(9ms):
1 class Solution { 2 public: 3 vector nextGreaterElement(vector & findNums, vector & nums) { 4 stack s ; 5 unordered_mapm ; 6 vector res ; 7 for(int n : nums){ 8 while(!s.empty() && n > s.top()){ 9 m[s.top()] = n ;10 s.pop() ;11 }12 s.push(n) ;13 }14 for(int i : findNums){15 if (m.count(i))16 res.push_back(m[i]) ;17 else18 res.push_back(-1) ;19 }20 return res ;21 }22 };