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.
idea:
单调递减栈➕hashmap
栈存nums2里的递减元素，在当前元素 > peek 的时候，pop元素出栈，map中存popo出来的元素与当前元素的pair，就是next greater element。如果栈为空或者当前元素 < peek，停止操作。
关键code:
for (int i : nums2) {
if (s.isEmpty()) {
s.push(i);
} else if (i < s.peek()) {
s.push(i);
} else {
//i > s.peek()
while (!s.isEmpty() && s.peek() < i) {
int temp = s.pop();
map.put(temp, i);
}
s.push(i);
}
}
复杂度：o（n）
