496. Next Greater Element I

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.
You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.
For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.
Return an array _ans of length nums1.length such that ans[i] is the next greater element as described above._

Q：数组元素的下一个最大值。

Constraints:

• 1 <= nums1.length <= nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 10^4
• All integers in nums1 and nums2 are unique.
• All the integers of nums1 also appear in nums2.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.

解决思路1：暴力搜索，双循环遍历

解决思路2：单调栈+散列表

遍历nums2，每遇到一个新元素，判断栈顶元素与新元素

• 如果栈顶元素大，则新元素不可能是栈内元素的下一个较大值，新元素入栈；
• 如果新元素大，则栈顶元素的下一个最大值是新元素，记入map中， 弹出栈顶元素；继续判断，直至栈顶元素大于新元素，然后将新元素入栈 ；
• 最后栈内剩余元素都是无下一个较大值的，也即其下一个较大值是-1。

• 维护的栈从栈顶到栈底是单调不降的（按入栈的顺序是单调不增的），因此称为单调栈。

  代码：

  class Solution {
  public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        map<int, int> num2ge;
        stack<int> ss;
        for(int& x:nums2){
            while(!ss.empty() && ss.top()<x){
                num2ge[ss.top()] = x;
                ss.pop();
            }
            ss.push(x);
        }
        while(!ss.empty()){
            num2ge[ss.top()] = -1;
            ss.pop();
        }
  
        vector<int> result(nums1.size(), 0);
        for(int i=0; i<nums1.size(); i++){
            result[i] = num2ge[nums1[i]];
        }
  
        return result;
    }
  };
  

  运行效率评价：

  Runtime: 8 ms, faster than 63.90% of C++ online submissions for Next Greater Element I.
  Memory Usage: 8.9 MB, less than 56.43% of C++ online submissions for Next Greater Element I.