496. Next Greater Element I

最新推荐文章于 2021-10-26 01:11:44 发布

原创最新推荐文章于 2021-10-26 01:11:44 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 同时被 2 个专栏收录

21 篇文章

订阅专栏

C/C++

11 篇文章

订阅专栏

本文介绍了一个算法问题，即如何在两个数组中找到第一个更大的数。具体来说，对于数组nums1中的每一个元素，在数组nums2中找到它右侧的第一个比它大的元素，如果不存在则返回-1。文章提供了一个C++实现的示例。

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.

思路：num1是num2的子集，从num2中找到num1中指定元素的下一个较大的值并返回，若没有，则返回-1。循环获取nums1中的每个元素，并在nums2中查找返回小标迭代器，然后记录下位置按规则向后遍历寻找。

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        vector<int> res;
        for(auto &elem : findNums)
        {
            auto it=find(nums.begin(),nums.end(),elem);
            if(++it==nums.end())
            {
                res.push_back(-1);
            }
            else
            {
                for(;it!=nums.end();++it)
                {
                    if(*it>elem)
                    {    res.push_back(*it);
                        break;
                    }
                    
                }
                if(it==nums.end())
                    res.push_back(-1);
            }
        }
        return res;
    }
};