LeetCode496. Next Greater Element I

题目：

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.

解题过程：

首先由于nums1是nums2的子集（而且不重复），所以我们可以给nums2写一个类似next数组的算法，就是用map去存储这个数的下一个比它大的数，

即map[i] = j;  i是该数，j是下面比该数大的数，然后用栈去模拟计算的过程。

代码如下：

class Solution {public:vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {stack<int> sta;map<int, int> mmap;vector<int> result;for (int i : nums) {while (!sta.empty() && sta.top() < i) {mmap[sta.top()] = i;sta.pop();}sta.push(i);}for (auto i : findNums) {if (mmap.count(i)) {result.push_back(mmap[i]);}else result.push_back(-1);}return result;}};