[untitled] Li Kou 496 Next larger element I

nums1  Middle number  x  Of Next bigger element Refer to  x  stay  nums2 Corresponding position in On the right side Of first Than  x  Big elements .

Here are two for you There are no repeating elements Array of  nums1 and  nums2 , Subscript from 0 Start counting , among nums1  yes  nums2  Subset .

For each 0 <= i < nums1.length , Find satisfaction nums1[i] == nums2[j] The subscript j , And in nums2 determine nums2[j] Of Next bigger element . If there is no next larger element , Then the answer to this query is -1 .

Returns a length of  nums1.length Array of ans As the answer , Satisfy ans[i] As mentioned above Next bigger element .

Example 1：

Input ：nums1 = [4,1,2], nums2 = [1,3,4,2].
Output ：[-1,3,-1]
explain ：nums1 The next larger element of each value in is described below ：
- 4 , Mark... In bold and italics ,nums2 = [1,3,4,2]. There is no next bigger element , So the answer is -1 .
- 1 , Mark... In bold and italics ,nums2 = [1,3,4,2]. The next bigger element is 3 .
- 2 , Mark... In bold and italics ,nums2 = [1,3,4,2]. There is no next bigger element , So the answer is -1 .
Example 2：

Input ：nums1 = [2,4], nums2 = [1,2,3,4].
Output ：[3,-1]
explain ：nums1 The next larger element of each value in is described below ：
- 2 , Mark... In bold and italics ,nums2 = [1,2,3,4]. The next bigger element is 3 .
- 4 , Mark... In bold and italics ,nums2 = [1,2,3,4]. There is no next bigger element , So the answer is -1 .

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/next-greater-element-i
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###  Their thinking

Here is the solution

###  Code

```javascript

/**

 * @param {number[]} nums1

 * @param {number[]} nums2

 * @return {number[]}

 */

var nextGreaterElement = function(nums1, nums2) {

   let a=[],flag=false,flagB=false

   for(let i=0;i<nums1.length;i++){

       for(let j=0;j<nums2.length;j++){

           if(nums1[i]==nums2[j]){

             

             flag=true

             continue

           }

           if(flag&&nums2[j]>nums1[i]){

                a.push(nums2[j])

                flagB=true

                break

           }

       }

       if(!flagB){a.push(-1)}

       flagB=false

       flag=false

   }

   return a

};

```