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  1. templates

Maxwindow

对一个数组,保持一个滑动窗口,求出每个窗口内的最大值。使用一个从左到右由大到小的队列,每次从队列前端取的数就是窗口的最大值。 如果新进来的数破坏了大小,则不断从队尾弹出元素,直至保持了由大到小的属性(相等时也需要弹出)。

struct WindowMax
{
    //求窗口的最大值,需要对队列维护一个信息,由大到小
    //遇到相同元素取后进的,因为虽然一样大,但是后进的晚过期
    int L{ -1 }, R{-1};//窗口的左右边界,R是窗口右边界的下一个位置
    //L和R分别窗口左右的上一个和下一个元素
    vector<int> nums;
    deque<int> Max;//存放元素的下标
    WindowMax(vector<int> n):nums(n),L(-1),R(0){}

    //窗口增大
    void AddFromRight() {
        if (R == nums.size()) {
            return;
        }
        while (!Max.empty() && Max.back() <= nums[R]) {
            Max.pop_back();
        }
        Max.push_back(R);
        R++;
    }
    //窗口减小,如果窗口左边界跨过了最大值队列,最大值队列的相对应小标弹出
    void RemoveFromLeft() {
        if (L > R - 1) {//L>=R
            return;
        }
        L++;
        //L是窗口左边界的上一个位置
        if (L == Max.front()) {
            Max.pop_front();
        }
    }
    int GetMax() {
        if (!Max.empty()) {
            return nums[Max.front()];
        }
        return INT_MIN;
    }

};

``` c++ "指定窗口大小时,求每个窗口的最大值" vector getMax(vector& nums, int k) { if (nums.empty() || knums.size()) { return vector {}; } //窗口大小为k deque Max; vector result(nums.size() - k + 1); int index{ 0 }; for (auto i = 0; i < (int)nums.size(); ++i) { //处理每个元素,如果破坏了队列的大小结构 while (!Max.empty() && Max.back() <= nums[i]) { //从队尾弹出元素 Max.pop_back(); } Max.push_back(i); //来到一个新位置,需要处理过期窗口的过期元素,处理队列中的过期最大值 if (Max.front() == i - k) { Max.pop_front(); } //窗口形成后,每一轮都会产生一个最大值 if (i >= k - 1) { result[index++] = nums[Max.front()]; } } return std::move(result);

}

```

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Last updated 4 years ago

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