496. Next Greater Element I（Stack-Easy）

转载请注明作者和出处：http://blog.csdn.net/c406495762

    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.

题目： 给定两个数组nums1和nums2，nums1是nums2的一个子集。找到所有nums1中的元素对应nums2位置的下一个更大的元素。如果有这个更大的元素，返回这个元素，如果没有返回-1。

思路：

1.挑出nums2中后一个元素比前一个元素大的一对元素写入关联容器unordered_map中，进行记录；

2.查看nums1中的元素是否存在于unordered_map中，如果有，取出这个值，如果没有返回-1。

Language : cpp

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        stack<int> s;
        unordered_map<int, int> m;
        //遍历nums中的元素
        for (auto e : nums){
            //堆栈s为空并且堆栈s的栈顶的元素小于e元素,将元素写入map中,key值为栈顶元素,value值为比栈顶元素大的元素
            while(!s.empty() && s.top() < e){
                m[s.top()] = e;
                s.pop();
            }
            s.push(e);
        }
        vector<int> ans;
        //查找findNums中元素,如果m中存在n,返回m[n]的valuse值,如果不存在返回-1
        for (auto n : findNums){
            ans.push_back(m.count(n) ? m[n] : -1);
        }
        return ans;
    }
};