The maximum value of the sliding window - cattle-off

Title Description
Given an array and the size of the sliding window, sliding window to find the maximum value among all values. For example, if the input array size and {2,3,4,2,6,2,5,1} 3 sliding window, then the presence of a total of six sliding window, their maximum values is {4,4,6, 6,6,5}; it has the following six array {2,3,4,2,6,2,5,1} for the sliding window: {[2,3,4], 2,6,2,5 , 1}, {2, [3,4,2], 6,2,5,1}, {2,3, [4,2,6], 2, 5}, {2,3,4 , [2,6,2], 5,1}, {2,3,4,2, [6,2,5], 1}, {2,3,4,2,6, [2,5, 1]}.
Solution:
1, define two vector arrays, array size elements stored at each sliding window, the maximum value stored MaxNum
2, (), array.end () array.begin () Get the maximum max_element by *
. 3, by vector :: iterator idx = array.begin () ; array.erase (idx); delete array element head

class Solution {
public:
    vector<int> maxInWindows(const vector<int>& num, unsigned int size)
    {
        vector<int> maxnum;//保存最大值
        vector<int> array;//保存每次窗口滑动后的数组
        unsigned int length = num.size();//原num数组的长度
        unsigned int k = 0;//记录遍历num的位置
        
        if (size>length || size<1)
            return maxnum;
        
        while (k < size-1)
            array.push_back(num[k++]);
        
        for (int i=0; i<length-size+1; i++)
        {
            array.push_back(num[k++]);
            maxnum.push_back(*max_element(array.begin(), array.end()));//保存当前滑动窗口内的最大值
            vector<int>::iterator idx = array.begin();
            array.erase(idx);//删除array中头部元素
        }
        return maxnum;
    }
};