【leetcode】496. Next Greater Element I

原題

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

查找下一個較大的元素
給出num1，num2,1是2的子集但亂序，找出1中元素在2中右側第一個較大的元素，若不存在則取-1

思路

個人思路很簡單，就是遍歷，但效率較低，找到一種比較清奇的思路：利用棧來裝num2的元素，若是新元素比棧頂的小，就繼續裝，若是比它大，就將棧頂元素和即將裝入的新元素成對裝在一個map中待用，一直裝map只到要入棧的元素比棧頂的小，再繼續
待所有裝完後，map中的元素對就都是該元素和他右側第一個較大的元素對了，此時只須要遍歷num1，從map中取值便可

個人解法

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;
    }

較優解

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;
    }