leetcode 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:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

想不到什么O(n)的好办法，只能对于num1中的每个整数，去nums2中一个个的找next更大的元素。

public class Next_Greater_Element_I_496 {

	public int findIt(int[] nums,int num){
		boolean isFind=false;
		for(int i=0;inum){
					return nums[i];
				}
			}
		}
		return -1;
	}
	
	public int[] nextGreaterElement(int[] findNums, int[] nums) {
		int[] result=new int[findNums.length];
		for(int i=0;i

看了大神的O(n)解答，我是膜拜的。用栈！！思路是先找到nums中每个元素的之后第一个比它大的元素，存入map。用栈的思想，将nums2里的数字不断地push进去，当push一个数字a时，判断a是否比栈顶的元素大，如果大，就pop栈顶元素，然后将map中填入(栈顶元素，a），再继续看栈顶元素，直到栈顶元素比a小为止，再把a push进栈。（一直留在栈里没有被pop出来的元素就是结果为-1的，后面没有比它大的了）
We use a stack to keep a  decreasing  sub-sequence, whenever we see a number  x  greater than  stack.peek()  we pop all elements less than  x  and for all the popped ones, their next greater element is  x
For example  [9, 8, 7, 3, 2, 1, 6]
The stack will first contain  [9, 8, 7, 3, 2, 1]  and then we see  6  which is greater than  1  so we pop  1 2 3  whose next greater element should be  6

public int[] nextGreaterElement(int[] findNums, int[] nums) {
        Map map = new HashMap<>(); 
        Stack 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;
    }

Nice solution！