Next Greater Element I

Stack, Hash Map

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

Analysis

Similar idea to Daily Temperatures problem.

Keep a monotonous stack. 经典问题,单调栈可以轻松解决。

Solution

Monotonous Stack + HashMap (4 ms, faster than 85.00%)

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap<>();
        Deque<Integer> stack = new ArrayDeque<>();
        int[] res = new int[nums1.length];

        for (int i = 0; i < nums2.length; i++) {
            while (!stack.isEmpty() && nums2[i] > nums2[stack.peek()]) {
                int idx = stack.pop();
                map.put(nums2[idx], nums2[i]);
            }
            stack.push(i);
        }

        for (int j = 0; j < nums1.length; j++) {
            res[j] = map.getOrDefault(nums1[j], -1);
        }

        return res;
    }
}

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