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:
nums1andnums2are unique.nums1andnums2would not exceed 1000.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; } }
原文：http://www.cnblogs.com/wxshi/p/7598550.html