leetcode-496-easy

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:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 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.
Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 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.
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?

思路一：用 map 存储 nums2 的值和下标，然后遍历 nums1 求解

    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int[] result = new int[nums1.length];
        Arrays.fill(result, -1);

        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums2.length; i++) {
            map.put(nums2[i], i);
        }

        for (int j = 0; j < nums1.length; j++) {
            int num = nums1[j];
            for (int i = map.get(num) + 1; i < nums2.length; i++) {
                if (nums2[i] > num) {
                    result[j] = nums2[i];
                    break;
                }
            }
        }

        return result;
    }

思路二：看了题解，发现有更好的解法，用单调栈求解

for(int i = nums2.length - 1; i >= 0; i--){
    while(!s.empty() && s.peek() <= nums2[i]){//栈顶只要小于待入栈元素就不断出栈，循环退出条件是栈顶元素大于nums2[i]，这样在入栈之后才能保持单调栈的情况
        s.pop();//持续出栈，构造单调栈
    }
    map.put(nums2[i],s.empty() ? -1 : s.peek());//map中记录对应的映射
    s.push(nums2[i]);//入栈
}