[leetcode-496-Next Greater Element I]

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.

思路:

使用一个栈，将nums中的数字依次进行入栈操作，在入栈之前，随时将要入栈的元素nums[i]与栈顶元素
进行比较，如果栈顶元素大于nums[i],则将nums[i]入栈，如果栈顶元素小于nums[i]，则说明栈顶元素的下一个最大的元素即为
nums[i],然后用一个map记录这个关系，然后将栈顶元素出栈

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums)
     {
         stack<int> st;
         map<int, int>M;
         for (int n : nums)
         {
             while (!st.empty() && st.top() < n)
             {
                 M[st.top()] = n;
                 st.pop();
             }
             st.push(n);
         }
         vector<int>ret;
         for (int n : findNums) ret.push_back(M.count(n) ? M[n] : -1);
         return ret;
     }

参考：

http://www.liuchuo.net/archives/3197

https://discuss.leetcode.com/topic/78397/c-stack-unordered_map