LeetCode 496. Next Greater Element I

496. Next Greater Element I

Total Accepted: 6642
Total Submissions: 11247
Difficulty: Easy
Contributors: love_FDU_llp

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.

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【题目分析】

给定两个数组num1和num2，两个数组中的元素都是唯一的并且num1是num2的一个子集。求num1中的元素对应的num2中相同的元素在num2中右边第一个大于它的元素。

【思路】

一开始就没有什么想法。大神的回答如下：

Key observation:
Suppose we have a decreasing sequence followed by a greater number
For example [5, 4, 3, 2, 1, 6] then the greater number 6 is the next greater element for all previous numbers in the sequence

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。

【java代码】

 1 public int[] nextGreaterElement(int[] findNums, int[] nums) {
 2         Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x
 3         Stack<Integer> stack = new Stack<>();
 4         for (int num : nums) {
 5             while (!stack.isEmpty() && stack.peek() < num)
 6                 map.put(stack.pop(), num);
 7             stack.push(num);
 8         }   
 9         for (int i = 0; i < findNums.length; i++)
10             findNums[i] = map.getOrDefault(findNums[i], -1);
11         return findNums;
12     }