496. Next Greater Element I
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package LeetCode_496 /** * 496. Next Greater Element I * https://leetcode.com/problems/next-greater-element-i/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: All elements in nums1 and nums2 are unique. The length of both nums1 and nums2 would not exceed 1000. * */ class Solution { fun nextGreaterElement(nums1: IntArray, nums2: IntArray): IntArray { //method 1: brute force val result = IntArray(nums1.size) for (i in nums1.indices) { var j = 0 while (j < nums2.size) { //find out the index of same digit if (nums1[i] == nums2[j]) { break } j++ } //scan nums2 from the index we find below to end //num1 = [2,4], nums2 = [1,2,3,4]. var k = j + 1 while (k < nums2.size) { //for example in [1,2,3,4], if 3>2 set to the result if (nums2[k] > nums2[j]) { result.set(i, nums2[k]) break } k++ } if (k == nums2.size) { result.set(i, -1) } } return result } }
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