496. Next Greater Element I

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本文介绍了一个算法问题，即如何在两个不重复元素的数组中找到第一个比特定元素大的数。对于数组nums1中的每个元素，在nums2中找到其右侧的第一个更大的元素。如果不存在，则返回-1。文章通过实例说明了这一过程，并提供了一段C++实现代码。

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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.

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        vector<int> res;
       int flag=0;
       for(int i=0;i<findNums.size();i++){
            flag=0;
           for(int j=0;j<nums.size();j++){
               if(findNums[i]==nums[j]){
                   for(int k=j+1;k<nums.size();k++){
                       if(nums[k]>nums[j]){
                           res.push_back(nums[k]);
                           flag=1;
                            break;
                       }
                   }
               }//endif
               if(flag==1) break;
           }
           if(flag==0) res.push_back(-1);
       }
       return res;
    }
};