23.leetCode496：Next Greater Element I

题目：
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.

题意：给定两个数组nums1和nums2，nums1是nums2的子集，nums1中的每个元素能在nums2中找到（位置可能不一样）。对于nums1中的每个元素e，找到nums2中的e之后第一个比e大的元素。

思路：将nums2中的元素放到map中，并与其下标映射起来。对于nums1中的每个元素e，首先找到nums2中e所在的下标index，然后从index+1开始往后找，找到比e大的元素就返回。若找不到比e大的元素，就返回-1.

代码：

class Solution {    public int[] nextGreaterElement(int[] nums1, int[] nums2) {        Map<Integer,Integer> map = new HashMap<Integer,Integer>();        int length = nums1.length;        int[] result = new int[length];        for(int i=0;i<nums2.length;i++){            map.put(nums2[i],i);        }        for(int j=0;j<length;j++){            int num = map.get(nums1[j]);            if(num==nums2.length-1)                result[j]=-1;            for(int k=num+1;k<nums2.length;k++){                if(nums2[k]>nums1[j]){                   result[j]=nums2[k];                    break;                }                result[j]=-1;            }        }        return result;    }}