496 Next Greater Element I
1. Question
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.
2. Implementation
思路: 题目已经说明nums1是nums2的子集,所以可以对nums2进行遍历,维护一个单调递减stack, 当栈顶的数小于当前nums2的数时,则nums2是栈顶数的Next Greater Element。同时通过一个HashMap记录每个元素的Next Greater Element即可
class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
Map<Integer, Integer> map = new HashMap<>();
Stack<Integer> stack = new Stack<>();
for (int num : nums2) {
while (!stack.isEmpty() && stack.peek() < num) {
map.put(stack.pop(), num);
}
stack.push(num);
}
int[] res = new int[nums1.length];
for (int i = 0; i < nums1.length; i++) {
res[i] = map.getOrDefault(nums1[i], -1);
}
return res;
}
}
3. Time & Space Complexity
时间和空间复杂度都为O(n), n为nums2里的元素个数
Last updated
Was this helpful?