leetcode 496. Next Greater Element I

川长思鸟来 2022-06-03 00:35 181阅读 0赞

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.

本题题意很简单，就是寻找在原数组中右边比自己大的元素，假如不存在返回-1，那么直接循环暴力求解即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <cmath>
    
    using namespace std;
    
    class Solution 
    {
    public:
        vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums)
        {
            vector<int> res;
            for (int a : findNums)
            {
                int i = 0;
                for (i = 0; i < nums.size(); i++)
                {
                    if (nums[i] == a)
                        break;
                }
                for (i=i+1; i < nums.size(); i++)
                {
                    if (nums[i] > a)
                        break;
                }
                res.push_back(i >= nums.size() ? -1 : nums[i]);
            }
            return res;
        }
    };