leetcode496 Next Greater Element JAVA

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Description

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.

解法1

public int[] nextGreaterElement(int[] findNums, int[] nums) {int[] res = new int[findNums.length];for(int i=0; i<findNums.length; i++) {int num1 = findNums[i];boolean flagNum1IsFound = false;boolean flagGreaterIsFound = false;for(int j=0; j<nums.length; j++) {if(nums[j] == num1) flagNum1IsFound = true;if(flagNum1IsFound && nums[j]>num1) {res[i] = nums[j];flagGreaterIsFound = true;break;}}if(!flagGreaterIsFound) res[i] = -1;}return res;}

解法2

利用栈的“后进先出”原则。先把num2数组中的每一个数的Greater Element都找到，存到Map中。最后num1中的数到Map中去对应，如果Map中存在，则返回其value,如果不存在，返回默认值-1。

public static int[] nextGreaterElement1(int[] findNums, int[] nums) {//Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of xStack<Integer> 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;}