[LeetCode]496. Next Greater Element I(下一个更大的元素 1)

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本文介绍了一个算法问题——寻找下一个更大的元素。给定两个数组nums1和nums2，其中nums1为nums2的子集，任务是找出nums1中每个元素在nums2中对应位置右侧的第一个更大元素。若不存在，则输出-1。

496. Next Greater Element I

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.
题目大意：
给出两个数组nums1 和 nums2，其中nums1 是 nums2 的子集。找出nums1的每个元素对应的The Next Greater Number.
nums1 中 x 的The Next Greater Number是指nums2中相对应数右边第一个比它大的元素，如果没有则返回-1。

举例1. 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.

举例2. 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.（所有元素都不同）
The length of both nums1 and nums2 would not exceed 1000.（长度小于1000）

C++

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        vector<int> A;
        for(int findNum : findNums)
        {
            bool B = false;
            for(int i=0;i<nums.size();i++)
            {
                if(findNum == nums[i])
                    B=true;
                if(B && findNum < nums[i])
                {
                    A.push_back(nums[i]);
                    B=false;
                    break;
                }
            }
            if(B)
                A.push_back(-1);
        }
        return A;
    }
};