LeetCode #496 - Next Greater Element I

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本文介绍了一个算法问题，即给定两个没有重复元素的数组，其中一个数组是另一个数组的子集。任务是找到第一个大于子数组中每个元素的对应元素。通过使用哈希表存储元素及其在原数组中的位置，实现了一种解决方案。

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题目描述：

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.

一个数组数另一个数组的子数组，那么对于子数组中的每个元素，求它在原数组的下一个更大的元素，保证数组中元素不重复。每次搜索是从某个元素在原数组的下标开始，但是对于子数组元素我们不知道它在原数组的下标，由于不存在重复元素，我们直接用一个哈希表来存储元素和它在原数组的下标。

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        unordered_map<int,int> hash;
        for(int i=0;i<nums.size();i++) hash[nums[i]]=i;
        vector<int> result(findNums.size(),-1);
        for(int i=0;i<findNums.size();i++)
        {
            for(int j=hash[findNums[i]]+1;j<nums.size();j++)
            {
                if(nums[j]>findNums[i])
                {
                    result[i]=nums[j];
                    break;
                }
            }
        }
        return result;
    }
};