# 496. Next Greater Element I

**496.Next Greater Element I**

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 array* `ans` *of length* `nums1.length` *such that* `ans[i]` *is the **next greater element** as described above.*

**Example 1:**

<pre><code><strong>Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
</strong><strong>Output: [-1,3,-1]
</strong><strong>Explanation: The next greater element for each value of nums1 is as follows:
</strong>- 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.
</code></pre>

**Example 2:**

<pre><code><strong>Input: nums1 = [2,4], nums2 = [1,2,3,4]
</strong><strong>Output: [3,-1]
</strong><strong>Explanation: The next greater element for each value of nums1 is as follows:
</strong>- 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.
</code></pre>

**Constraints:**

* `1 <= nums1.length <= nums2.length <= 1000`
* `0 <= nums1[i], nums2[i] <= 104`
* 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?

***My Solutions:***

这道题的意思是， 对于nums1里的每一个数，求在nums2里比TA第一个大的数。

Key observation:&#x20;

Suppose we have a decreasing sequence followed by a greater number, for example \[5, 4, 3, 2, 1, 6]. The greater number 6 is the next greater element for all previous numbers in the sequence.

Use a stack to keep a decreasing sub-sequence. Whenever we see a number x greater than stack.peek(), 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. We pop 1 2 3 whose next greater element should be 6

先遍历nums2，建立一个递减的stack。如果当前数i比stack.peek()小，就入栈；不然就出栈所有比当前i小的元素，把小数value和next greater element = i 记录在map里。

```java
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
    // 用来储存数字和他们的第一大值的index
    Map<Integer, Integer> map = new HashMap<>();
    Stack<Integer> stack = new Stack<>();
    for (int i : nums2) {
        // pop the numbers smaller than the current number out of the stack
        while (!stack.isEmpty() && stack.peek() < i) {
            // put (small number value -> the next great element of this small number) to the map
            map.put(stack.pop(), i); 
        }
        stack.push(i);
    }
    
    int[] res = new int[nums1.length];
    for (int i = 0; i < nums1.length; i++) {
        // 如果map里没有nums1当前的数，说明nums2里没有比当前数更大的数，所以记为-1
        res[i] = map.getOrDefault(nums1[i], -1);
    }
    return res;
}
```