Next Greater Element I
Stack
, Hash Map
You are given two arrays (without duplicates)nums1
andnums2
wherenums1
’s elements are subset ofnums2
. Find all the next greater numbers fornums1
's elements in the corresponding places ofnums2
.
The Next Greater Number of a numberxinnums1
is the first greater number to its right innums2
. 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.
Analysis
Similar idea to Daily Temperatures problem.
Keep a monotonous stack. 经典问题,单调栈可以轻松解决。
Solution
Monotonous Stack + HashMap (4 ms, faster than 85.00%)
class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
Map<Integer, Integer> map = new HashMap<>();
Deque<Integer> stack = new ArrayDeque<>();
int[] res = new int[nums1.length];
for (int i = 0; i < nums2.length; i++) {
while (!stack.isEmpty() && nums2[i] > nums2[stack.peek()]) {
int idx = stack.pop();
map.put(nums2[idx], nums2[i]);
}
stack.push(i);
}
for (int j = 0; j < nums1.length; j++) {
res[j] = map.getOrDefault(nums1[j], -1);
}
return res;
}
}
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