leetcode -- 496. Next Greater Element I 【栈】

题目

You are given two arrays (without duplicates) nums1 andnums2 wherenums1’s elements are subset ofnums2. Find all the next greater numbers fornums1's elements in the corresponding places ofnums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right innums2. 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.

题意：

给定两个数组（没有重复）nums1 和 nums2 ,nums1 是nums2 的子集。找到nums1中每个元素的在nums2中的下一个比其大的值。

nums1中的x的下一个大的值是该数字x在nums2中，右边的比其大的一个值。如果不存在，返回-1。

THINKING :

1. 肯定不能引入排序，因为排序会将次序打乱，进而破坏问题的条件前提。

2.局部就近匹配，符合栈的特点。（similar to 括号匹配问题）

题目解答

方法1（简洁，技巧）：

• 【 栈 】利用栈来为nums2中的每个数字分配下一个最大值。（栈确实能达到此类效果）
• 【 栈的特性 】正常循环的情况下，数组的滚动（游标移动）是向后的，引入栈的时候，则可以有了向前滚动的机会（有了一定的反悔的机会），然后这样子就能够解决一些局部的问题（比如说，寻找相邻的大的数字）。由于栈还可以对于没有价值（已经发现了大的数字）的东西删除，这样子的遗忘功能，简化了搜索空间，问题空间。
  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

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

方法2（复杂，一般）：

 public int[] nextGreaterElement(int[] findNums, int[] nums) {

		int results[] = new int[findNums.length];
		for (int j = 0; j < findNums.length; j++) {
			boolean findBigger = false;
			boolean isRight = false;
			
			for (int i = 0; i < nums.length; i++) {
				
				if (nums[i] == findNums[j]) {
					isRight = true;
				}
				
				if (isRight && findNums[j] < nums[i]) {
					results[j] = nums[i];
					findBigger = true;
					break;
				}
			}
			
			if (!findBigger) {
				results[j] = -1;
			}
		}
		return results;
	}