Description

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

1.All elements in nums1 and nums2 are unique.
2.The length of both nums1 and nums2 would not exceed 1000.

Solution

这道题目和739DailyTemperature有相似之处，739的栈解法很惊艳，自己试了一下用栈解决，可是由于对栈不太熟悉的原因，写起来很吃力。于是，我想起来LeetCode1用过的哈希方法，而且键和索引反着写，适合搜索存储nums1的元素在nums2的索引，非常合适。

Code

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int l=nums1.length;
        //int data[]=new int[nums1.length];
        int data1[]=new int[nums1.length];
        int m=0;
        //int n=0;
        
       Map<Integer,Integer>map=new HashMap<Integer,Integer>();
        for(int i=0;i<nums2.length;i++){
            map.put(nums2[i],i);
        }
        for(int i=0;i<l;i++){
            data1[i]=-1;
            int k=map.get(nums1[i]);
            
            for(int j=k+1;j<nums2.length;j++){
                if(nums2[j]>nums1[i]) {data1[i]=nums2[j];
                //else data1[m++]=-1;
                break;
                                      }
            }
        }
    return data1;
    }
}

遇到的问题

1.Line14：直接先行赋值-1比较简便，减少比较的过程
2.Line20：一定要break，不然非-1出输出的值可能不对