496. 下一个更大元素 I

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题目

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 arrayans of lengthnums1.length such thatans[i] is the next greater element as described above.

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.

Constraints:

• 1 <= nums1.length <= nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 10^4
• 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?

题目大意

题目给出 2 个数组 A 和 B，针对 A 中的每个数组中的元素，要求在 B 数组中找出比 A 数组中元素大的数，B 中元素之间的顺序保持不变。如果找到了就输出这个值，如果找不到就输出 -1。

解题思路

• 使用单调递增栈；
• 因为 nums1 是 nums2 的子集，所以我们可以先遍历一遍 nums2，并构造单调递增栈；
• 求出 nums2 中每个元素右侧下一个更大的元素，然后将其存储到哈希表中；
• 再遍历一遍 nums1，从哈希表中取出对应结果，存放到答案数组中；
• 这种解法的时间复杂度是 O(n)。

代码

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var nextGreaterElement = function (nums1, nums2) {
	let map = new Map();
	let stack = [];
	for (let num of nums2) {
		while (stack.length && stack[stack.length - 1] < num) {
			map.set(stack.pop(), num);
		}
		stack.push(num);
	}
	for (let i = 0; i < nums1.length; i++) {
		nums1[i] = map.has(nums1[i]) ? map.get(nums1[i]) : -1;
	}
	return nums1;
};

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