August's Box

Next Greater Element I

Description

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 and nums2 are unique.
  • The length of both nums1 and nums2 would not exceed 1000.

Solution

题目的意思是:
找出数组中某一些数后面第一个大于这个数的数,不存在为-1

解法:
借助stack和map,通过比较、入栈和出栈更新数后面第一个大于这个数的数。代码如下:

python
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def nextGreaterElement(self, findNums, nums):
"""
:type findNums: List[int]
:type nums: List[int]
:rtype: List[int]
"""
mp = collections.defaultdict(int)
s, result = [], []
for n in nums:
while len(s) and s[-1] < n:
mp[s[-1]] = n
s.pop()
s.append(n)
for n in findNums:
result.append(mp.get(n, -1))
return result

c++
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vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
stack<int> s;
unordered_map<int, int> mp;
vector<int> result;
for(auto n : nums) {
while(!s.empty() && s.top() < n) {
mp[s.top()] = n;
s.pop();
}
s.push(n);
}
for(auto n : findNums)
result.push_back(mp.count(n) ? mp[n] : -1);
return result;
}