[LeetCode] 496. Next Greater Element I

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.

下一个更大元素I。

题意是给定两个没有重复元素的数组 nums1 和 nums2 ，其中nums1 是 nums2 的子集。找到 nums1 中每个元素在 nums2 中的下一个比其大的值。nums1 中数字 x 的下一个更大元素是指 x 在 nums2 中对应位置的右边的第一个比 x 大的元素。如果不存在，对应位置输出-1。

思路是单调栈。创建一个hashmap，记录num2中每个元素和他们各自的下一个更大元素。再遍历num1，看num1里面的元素是否在hashmap中有对应的value，有就输出，没有就返回-1。

时间O(n)

空间O(n)

Java实现

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        HashMap<Integer, Integer> map = new HashMap<>();
        Stack<Integer> stack = new Stack<>();
        for (int num : nums2) {
            while (!stack.isEmpty() && stack.peek() < num) {
                map.put(stack.pop(), num);
            }
            stack.push(num);
        }
        int[] res = new int[nums1.length];
        for (int i = 0; i < res.length; i++) {
            res[i] = map.getOrDefault(nums1[i], -1);
        }
        return res;
    }
}

JavaScript实现

/**
 * @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 > 0 && stack[stack.length - 1] < num) {
            map.set(stack.pop(), num);
        }
        stack.push(num);
    }
    let res = new Array(nums1.length).fill(-1);
    for (let i = 0; i < res.length; i++) {
        res[i] = map.get(nums1[i]) || -1;
    }
    return res;
};

LeetCode 题目总结