Next Greater Element I个人解决方法

题目来源LeetCode

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]

解题思路：先建立indexs来储存子数组中每个元素在nums2中的下标，然后再按照子数组中每个元素在nums2的下标来往后找比这个数大的元素，找得到就储存在results中，找不到就储存-1，具体代码如下：

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        int size1 = findNums.size();
        int size2 = nums.size();  
        vector<int>results;
        vector<int>indexs;
        for(int i = 0; i < size1; i++){
            for(int j = 0 ; j < size2; j++){
              if(findNums[i] == nums[j]){
                indexs.push_back(j);
                break;
              }
            }
        }
        for(int i = 0; i < size1; i++){
            for(int j = index[i]; j < size2; j++){
                if(nums[j] > findNums[i]){
                    results.push_back(nums[j]);
                    break;
                }
                else if(j == size2-1 && nums[j] <= findNums[i]){
                    results.push_back(-1);
                    break;
                }
            }
        }
        return results;
    }
};