leetcode 503. Next Greater Element II

最新推荐文章于 2025-01-28 06:45:00 发布

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最新推荐文章于 2025-01-28 06:45:00 发布

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本文介绍了一种解决循环数组中寻找每个元素下一个更大元素问题的高效算法，并提供了详细的C++实现代码。通过栈的数据结构优化查找过程，确保了算法的时间效率。

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题目：

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.

class Solution {
public:
    vector<int> nextGreaterElements(vector<int>& nums) {
        vector<int> fd(nums.size()*2,0);
        stack<pair<int,int>> re;
        nums.insert(nums.end(), nums.begin(), nums.end());
        for(int i = 0; i < nums.size();i++) {
            if(re.empty()) {
                re.push(pair<int,int>(nums[i],i));
            } else {
                if(nums[i] < re.top().first) {
                    re.push(pair<int,int>(nums[i],i));
                } else {
                    while(!re.empty()&&nums[i] > re.top().first) {
                        fd[re.top().second] = nums[i];
                        re.pop();
                    }
                    re.push(pair<int,int>(nums[i],i));
                }
            }
        }
        while(!re.empty()) {
            fd[re.top().second] = -1;
            re.pop();
        }
        fd.resize(fd.size()/2);
        return fd;
    }
};