leetcode-标签为stack 496. Next Greater Element I

原题

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.

代码分析

求一个元素的下一个更大值，比如[3 1 2] ，比 3 大的下一个更大值不存在，比 1 大的下一个更大值为 2，比2大的更大值不存在。

给定数组 [9 8 7 2 1 5].如何求出这个序列中每个元素的下一个最大值呢？

维护一个栈，使得序列为降序排序，比如当前的栈为 [9 8 7 2 1 ]，当元素 5 与栈顶元素 1 比较时，1 的下一个更大元素为 5，然后1出栈，同理 2 的更大值为 5，然后 2 出栈，不大于7，最后 5 入栈，此时的栈内元素序列为 [9 8 7 5]。

代码实现

上面的过程兑现为代码：

        // map from x to next greater element of x        private Dictionary<int, int> _dict = new Dictionary<int, int>();       private void preProcess(int[] nums)        {            Stack<int> stack = new Stack<int>();            foreach (var num in nums)            {                while (stack.Count > 0 && stack.Peek() < num)                    _dict.Add(stack.Pop(), num);                stack.Push(num);            }        }

现在给定一个nums的子序列 findnums，求得它其中每个元素的下一个更大值。

        public int[] NextGreaterElement(int[] findNums, int[] nums)        {            preProcess(nums);            for (int i = 0; i < findNums.Length; i++)            {                if (_dict.ContainsKey(findNums[i]))                    findNums[i] = _dict[findNums[i]];                else findNums[i] = -1;            }            return findNums;        }