496 Next Greater Element I

1. Question

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:
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Input:
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nums1 = [4,1,2],
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nums2 = [1,3,4,2].
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Output: [-1,3,-1]
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Explanation:
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For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
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For number 1 in the first array, the next greater number for it in the second array is 3.
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For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
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Example 2:
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Input:
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nums1 = [2,4],
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nums2 = [1,2,3,4].
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Output: [3,-1]
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Explanation:
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For number 2 in the first array, the next greater number for it in the second array is 3.
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For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
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Note:
  1. 1.
    All elements in nums1andnums2are unique.
  2. 2.
    The length of bothnums1andnums2would not exceed 1000.

2. Implementation

思路: 题目已经说明nums1是nums2的子集,所以可以对nums2进行遍历,维护一个单调递减stack, 当栈顶的数小于当前nums2的数时,则nums2是栈顶数的Next Greater Element。同时通过一个HashMap记录每个元素的Next Greater Element即可
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class Solution {
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public int[] nextGreaterElement(int[] nums1, int[] nums2) {
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Map<Integer, Integer> map = new HashMap<>();
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Stack<Integer> stack = new Stack<>();
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for (int num : nums2) {
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while (!stack.isEmpty() && stack.peek() < num) {
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map.put(stack.pop(), num);
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}
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stack.push(num);
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}
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int[] res = new int[nums1.length];
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for (int i = 0; i < nums1.length; i++) {
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res[i] = map.getOrDefault(nums1[i], -1);
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}
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return res;
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}
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}
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3. Time & Space Complexity

时间和空间复杂度都为O(n), n为nums2里的元素个数