【leetcode】496. Next Greater Element I

原题

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.

解析

查找下一个较大的元素
给出num1，num2,1是2的子集但乱序，找出1中元素在2中右侧第一个较大的元素，若不存在则取-1

思路

个人思路很简单，就是遍历，但效率较低，找到一种比较清奇的思路：利用栈来装num2的元素，若是新元素比栈顶的小，就继续装，若是比它大，就将栈顶元素和即将装入的新元素成对装在一个map中待用，一直装map只到要入栈的元素比栈顶的小，再继续
待所有装完后，map中的元素对就都是该元素和他右侧第一个较大的元素对了，此时只须要遍历num1，从map中取值便可

个人解法

public int[] nextGreaterElement(int[] findNums, int[] nums) {
        int[] result = new int[findNums.length];
        int n = 0;
        for (int i = 0; i < findNums.length; i++) {
            for (int j = 0; j < nums.length; j++) {
                if (nums[j] == findNums[i]) {
                    boolean isFind = false;
                    for (int k = j + 1; k < nums.length; k++) {
                        if (nums[k] > findNums[i]) {
                            result[n++] = nums[k];
                            isFind = true;
                            break;
                        }
                    }
                    if (!isFind) {
                        result[n++] = -1;
                    }
                    break;
                }
            }
        }
        return result;
    }

较优解

public int[] nextGreaterElementOptimize(int[] findNums, int[] nums) {
        Map<Integer, Integer> nextGreater = new HashMap<>();
        Stack<Integer> stack = new Stack<>();
        for (int num : nums) {
            while (!stack.isEmpty() && stack.peek() < num) {
                nextGreater.put(stack.pop(), num);
            }
            stack.push(num);
        }
        for (int i = 0; i < findNums.length; i++) {
            findNums[i] = nextGreater.getOrDefault(findNums[i], -1);
        }
        return findNums;
    }