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.

思路：

方法一：最笨的方法可以循环遍历两个数组，但是时间复杂的太高。

方法二：对第二个数组nums中的每个元素，找到它的Next Greater Element，即对应的下一个比它大的元素，形成<key,value>对并保存在HashMap中。然后遍历第一个数组findNums中每个元素，从HashMap中找到对应的Next Greater Element，存进数组对应位置。

这里用到了栈。【 栈 】正常循环的情况下，数组的滚动是向后的，引入栈后，则可以向前滚动，这样就能够解决一些局部的问题（比如寻找相邻的大的数字）。同时栈还可以删除没有价值的元素，具备遗忘功能。用到了栈的isEmpty()方法和HashMap的put()方法、getOrDefault()方法。

public class Solution {    public int[] nextGreaterElement(int[] findNums, int[] nums) {        Map<Integer,Integer> map = new HashMap<>();        Stack<Integer> stack = new Stack<>();        for(int num:nums){            while(!stack.isEmpty()&&stack.peek()<num){//用while而不能用if                map.put(stack.pop(),num);            }            stack.push(num);        }        for (int i=0;i<findNums.length;i++){            findNums[i]=map.getOrDefault(findNums[i],-1);        }        return findNums;    }}

程序的第6行应使用while判断并循环，不能使用if，因为存在下面这种情况，用if的结果是错误的。对于递减的元素，遇到一个大数时，需要把栈里的元素都依次弹出，来找到之前递减元素的对应Next Greater Element。Input:
[1,3,5,2,4]
[6,5,4,3,2,1,7]
Output:
[7,-1,-1,-1,-1]
Expected:
[7,7,7,7,7]