[LeetCode] Next Greater Element I

题目链接在此

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.

大概就是找出表2中所有数字的 Next Greater Element ：右侧比他大的第一个数。表一相当于是个是query list，查询单。用到栈和哈希。

顺便发现了个新大陆：for (int n : nums) 

C++ 11 可以这样写。（别鄙视我）

class Solution {public:vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {stack<int> s;unordered_map<int, int> m;for (int n : nums) {while(!s.empty() && s.top() < n) {m[s.top()] = n;  //  栈里面比n小的数，他们的Next Greater Element都是ns.pop();}s.push(n);}vector<int> ans;for (int n : findNums) {ans.push_back(m.count(n) != 0 ? m[n] : -1);}return ans;}};