/*****************************************************问题描述*************************************************
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.
给定两个整数数组，第一个数组的元素在第二个数组中都能找到，第一个数组中的各个元素，在第二个数组中先找到，然后在第二个数组中找到
的那个位置向右找，找到第一个比它大的元素返回，否则返回-1.
/*****************************************************我的解答*************************************************
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var nextGreaterElement = function(nums1, nums2) {
var retArray = [];
for(var index = 0; index < nums1.length; index++)
{
retArray.push(-1);
var temp = nums1[index];
var indexInNum2 = nums2.indexOf(temp);
for(var index2 = indexInNum2; index2 < nums2.length; index2++)
{
if(nums2[index2] > temp)
{
retArray[index] = nums2[index2];
break;
}
}
}
return retArray;
};
console.log(nextGreaterElement([2,4],[1,2,3,4]));
