503. Next Greater Element II

最新推荐文章于 2022-08-22 21:35:39 发布

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针对循环数组，寻找每个元素的下一个更大数值。若不存在，则输出-1。此问题需通过循环遍历解决。

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Description：

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.

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解题思路：

这题相对题I变化的是变成了一个循环数组，需要从数组尾部返回到数组头部，因此需要对第n取余。即遍历n个数，对于遍历的每个数字，再遍历其它数字，遇到较大值则break。

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