leetcode496. Next Greater Element I

496. Next Greater Element I

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.

解法一

在第二个数组中找到该值，然后从第二个数组的下一个值开始比较，如果大于则返回该数。

public class Solution {    public int[] nextGreaterElement(int[] findNums, int[] nums) {        if (findNums == null || nums == null || findNums.length > nums.length) {            return null;        }        int flag = 0;        int[] ret = new int[findNums.length];        for (int i = 0; i < findNums.length; i++) {            flag = 0;            ret[i] = -1;            for (int j = 0; j < nums.length; j++) {                if (nums[j] == findNums[i]) {                    flag = 1;                    continue;                }                if (flag == 1) {                    if (nums[j] > findNums[i]) {                        ret[i] = nums[j];                        flag = 0;                        break;                    }                }            }        }        return ret;    }}

解法二

利用栈，将第二个数组逐一入栈，如果栈顶元素小于数组的下一个数，则将栈顶元素和下一个数存到hashmap中，并将栈顶元素出栈。最后遍历第一个数组，到map中寻找是否存在该key，不存在则返回－1。

public class Solution {    public int[] nextGreaterElement(int[] findNums, int[] nums) {        if (findNums == null || nums == null || findNums.length > nums.length) {            return null;        }        Map<Integer, Integer> map = new HashMap<Integer, Integer>();        Stack<Integer> stack = new Stack<Integer>();        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.containsKey(findNums[i]) ? map.get(findNums[i]) : -1;            findNums[i] = map.get(findNums[i]) != null ? map.get(findNums[i]) : -1;        }        return findNums;    }}