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:
All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.

Solution1：HashMap

思路：
用哈希表先来建立每个数字和其坐标位置之间的映射，那么我们在遍历子集合中的数字时，就能直接定位到该数字在原数组中的位置，然后再往右边遍历寻找较大数即可

Time Complexity: O(N^2) Space Complexity: O(N)

Solution2：单调栈

思路：
For example [5, 4, 3, 2, 1, 6] then the greater number 6 is the next greater element for all previous numbers in the sequence

We use a stack to keep a decreasing sub-sequence, whenever we see a number x greater than stack.peek() we pop all elements less than x and for all the popped ones, their next greater element is x

For example [9, 8, 7, 3, 2, 1, 6]
The stack will first contain [9, 8, 7, 3, 2, 1] and then we see 6 which is greater than 1 so we pop 1 2 3 whose next greater element should be 6

Time Complexity: O(N) Space Complexity: O(N)

Solution1 Code：

class Solution {
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        Map m = new HashMap<>();
        // go through each element in nums and set its location in HashMap
        for(int i =0;ifindNums[i])
                {
                    minIndex =index;
                    break;
                }
            }
            if(minIndex ==-1) findNums[i] = -1;
            else findNums[i] = nums[minIndex];
        }
        return findNums;
    }
}

Solution2 Code：

class Solution {
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        Map map = new HashMap<>(); // map from x to next greater element of x
        Stack stack = new Stack<>();
        for (int num : nums) {
            while (!stack.isEmpty() && stack.peek() < num)
                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;
    }
}