496. Next Greater Element I

1. Question

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

2. Examples

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:

- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:

- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.

3. Constraints

  • 1 <= nums1.length <= nums2.length <= 1000
  • 0 <= nums1[i], nums2[i] <= 104
  • All integers in nums1 and nums2 are unique.
  • All the integers of nums1 also appear in nums2.

Follow up: Could you find an O(nums1.length + nums2.length) solution?

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/next-greater-element-i 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

5.1. 常规套路

class Solution {
  public int[] nextGreaterElement(int[] nums1, int[] nums2) {
    int m = nums1.length;
    int n = nums2.length;
    int[] res = new int[m];
    for (int i = 0; i < m; i++) {
      int j = 0;
      while (j < n && nums1[i] != nums2[j]) {
        j++;
      }
      int k = j + 1;

      while (k < n && nums2[k] <= nums2[j]) {
        k++;
      }

      res[i] = k < n ? nums2[k] : -1;
    }
    return res;
  }
}

5.2. 单调栈

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

    // 将nums2数组从后往前遍历,同时维护单调栈,数小弹栈,每次都要压进新值
    // 此处维护栈采用的是nums2的下标,容易混淆
    for (int i = nums2.length - 1; i >= 0; i--) {
      while (!stack.isEmpty() && nums2[stack.peek()] < nums2[i]) {
        stack.pop();
      }
      map.put(nums2[i], stack.isEmpty() ? -1 : nums2[stack.peek()]);
      stack.push(i);
    }

    for (int i = 0; i < nums1.length; i++) {
      res[i] = map.get(nums1[i]);
    }

    return res;
  }
}
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