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.

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关键在于理解题意，对于nums1中每一个数，在nums2中相同的数的位置的右边，找到第一个大于它的数。
第一种方法简单粗暴。O（n2）的复杂度，beats 16%。

public class Solution {    public int[] nextGreaterElement(int[] findNums, int[] nums) {        int[] output = new int[findNums.length];        //循环判断nums1的每个数        for(int i = 0; i < findNums.length; i++) {            //先预置找不到最大值，即-1            output[i] = -1;            //递减找到最近的great number，如果递减到两个数相等，说明相同数以右没有大于他的数            for(int j = nums.length - 1; j >= 0; j--) {                if(nums[j] == findNums[i]) break;                if(nums[j] > findNums[i]) output[i] = nums[j];            }        }        return output;    }}

后来Discuss中看到了一种相当巧妙的思路：Discuss链接，考虑到找greater number是从最右边开始的，可以利用stack和map来维护这个过程。从查询数组最右边开始循环，当栈头有比当前数小的值时则弹出，直到找到大于当前数的值，即是当前数的greater number，将这对数存到map中，并将当前数保存到栈内。可以发现，循环后的栈是查询数组的逆序保存，恰好和题目所要求的从右边找起的目的契合。通过这个算法时间复杂度一下干到了O(n)，膜拜=。=

public class Solution {    public int[] nextGreaterElement(int[] findNums, int[] nums) {        Map<Integer, Integer> map = new HashMap<>();        Stack<Integer> stack = new Stack<>();        for(int i = nums.length-1; i>=0; i--){            while(!stack.empty() && nums[i]>stack.peek()) stack.pop();            map.put(nums[i], (stack.empty())? -1 : stack.peek());            stack.push(nums[i]);        }        for(int i = 0; i<findNums.length; i++){            findNums[i] = map.get(findNums[i]);        }        return findNums;           }}