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  • Next Greater Element I
    알고리즘 2019. 7. 29. 17:41

    LeetCode 496번

    Tags. Stack

    https://leetcode.com/problems/next-greater-element-i/

     

    Next Greater Element I - LeetCode

    Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.

    leetcode.com

    문제

    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.

    Code

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    class Solution {
        public int[] nextGreaterElement(int[] nums1, int[] nums2) {
            
            int[] result = new int[nums1.length];
            
            for(int i = 0; i < nums1.length; i++) {
                boolean isit = false;
                result[i] = -1;
                for(int j = 0; j < nums2.length; j++) {
                    if(nums1[i] == nums2[j])
                        isit = true;
                    
                    if(nums1[i] < nums2[j] && isit) {
                        result[i] = nums2[j];
                        break;
                    }
                }
            }
            return result;
        }
    }
    cs

     

    Solution

    nums1 은 nums2의 부분집합이며 nums1의 각 원소가 nums2의 배열상 같은 원소의 오른쪽으로 큰 값이 있으면 결과배열에 넣어 반환하는 문제이다.

    스택을 사용하지 않고도 배열과 조건문으로 문제가 쉽게 풀 수 있었다.

     

    Runtime : 11ms

    Memory : 36.6MB

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