496. Next Greater Element I

Posted 2020-10-08 Wanna_Go

You are given two arrays(without duplicates)nums1andnums2wherenums1’s elements are subset ofnums2. Find all the next greater numbers fornums1‘s elements in the corresponding places ofnums2.

The Next Greater Number of a numberxinnums1is the first greater number to its right innums2. 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 innums1andnums2are unique.
2. The length of bothnums1andnums2would not exceed 1000.

题目的意思是nums1是nums2的子集，求nums1中的每一个元素在nums2中对应位置的右边第一个比它大的元素。没有则为-1，这一题读了半个小时没读懂=。=

例如Example1 中的1，在nums2中的右边元素为[3,4,2]，第一个比1大的是3。

暴力遍历法：

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int result[] = new int[nums1.length];
        
        for(int i = 0; i <nums1.length; i++)
        {
            int tmp =  Integer.MAX_VALUE;
           for(int j = 0; j<nums2.length; j++)
            {
               if(nums1[i] == nums2[j])
                    tmp = nums1[i];
               if(nums2[j] > tmp)
               {
                   result[i] = nums2[j];
                   break;
               }
                   
            } 
        }
         for(int i = 0;i < result.length;i++)
            if(result[i] == 0)result[i] = -1;
         return result;              
    }
}