LeetCode 496. Next Greater Element I 自我反思

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:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

这题就想到暴力搜索，但是太慢了。于是想着要创建一个map，保存一次遍历nums2的键值对，key为元素，value为下一个较大值；这里还要用一个栈，这个我想了半天，无法直接想出来这个栈，看了题解知道这个栈，就是保存之前遍历的值中还没有找到它的下一个较大值，所以感觉自己很LOW，慢慢学习吧。

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        stack<int> stack_index;
        unordered_map<int,int> mymap;
        for(int i = 0 ; i < nums.size() ; i ++)
        {
            while(!stack_index.empty() && nums[i] > nums[stack_index.top()])
            {
                mymap[nums[stack_index.top()]] = nums[i];
                stack_index.pop();
            }
            stack_index.push(i);
        }
        for(int i = 0 ; i < findNums.size() ; i++)
            findNums[i] = (mymap[findNums[i]] == 0)?-1:mymap[findNums[i]];
        return findNums;
    }
};