LeetCode #503 - Next Greater Element II

最新推荐文章于 2024-12-03 14:21:56 发布

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最新推荐文章于 2024-12-03 14:21:56 发布

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本文介绍了一种针对循环数组的NextGreaterElement算法实现。该算法寻找数组中每个元素的下一个更大元素，若不存在则输出-1。示例输入[1,2,1]，输出为[2,-1,2]。文章提供了完整的C++代码实现，适用于解决循环数组的NextGreaterElement问题。

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题目描述：

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.

Example 1:

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2; 
The number 2 can't find next greater number; 
The second 1's next greater number needs to search circularly, which is also 2.

Note: The length of given array won't exceed 10000.

和Next Greater Element类似，只不过数组变成了循环数组，所以搜索的步骤要做修改。

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