Next Greater Element I问题及解法

问题描述：

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.

示例：

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.

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.

问题分析：

题目说对第一个数组中的任何一个元素，都是第二个数组的元素，请找到在第二个数组中对应第一个数组元素的值的右边有没有比它还大的首个元素的值。

过程详见代码：

class Solution {
public:
    vector nextGreaterElement(vector& findNums, vector& nums) {
        vector res;
        map mapping;
        for(int i = 0;i < nums.size(); i++) mapping.insert(pair(nums[i],i));
        for(int i = 0;i < findNums.size();i++)
        {
        	int r = -1;
        	for(int j = mapping[findNums[i]] + 1; j < nums.size(); j++)
        	{
        		if(nums[j] > findNums[i])
        		{
        			r = nums[j];
        			break;
				}
			}
			res.push_back(r);
		}
		return res;
    }
};