Next Greater Element II

最新推荐文章于 2020-06-14 02:02:27 发布

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最新推荐文章于 2020-06-14 02:02:27 发布

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分类专栏： Leetcode 文章标签： C++

本文链接：https://blog.youkuaiyun.com/fujch3/article/details/60467157

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针对LeetCode中关于循环数组寻找每个元素的下一个更大数值的问题，本文介绍了一种通过复制数组来模拟循环特性的解决方案，并给出了具体的C++实现代码。

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题目来源LeetCode

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.

解题思路：将数组复制两遍来进行circular，然后再进行比较找出下一个最大元素，具体代码如下：

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