【LeetCode 239】 Sliding Window Maximum

題目描述

Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the max sliding window.

Follow up:

Could you solve it in linear time?

Example:

Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7] 
Explanation: 

Window position                Max
---------------               -----
[1  3  -1] -3  5  3  6  7       3
 1 [3  -1  -3] 5  3  6  7       3
 1  3 [-1  -3  5] 3  6  7       5
 1  3  -1 [-3  5  3] 6  7       5
 1  3  -1  -3 [5  3  6] 7       6
 1  3  -1  -3  5 [3  6  7]      7

Constraints:

1 <= nums.length <= 10^5
-10^4 <= nums[i] <= 10^4
1 <= k <= nums.length

思路

思路一：優先隊列，用一個vis數組標記左區間是否已經出隊列。
思路二：set

代碼

代碼一：

struct cmp {
    bool operator () (pair<int, int>& A, pair<int, int>& B) {
        return A.second < B.second;
    }
};

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        priority_queue<pair<int, int>, vector<pair<int, int>>, cmp> pq;
        int n = nums.size();
        vector<int> res;
        int cnt = 0;
        vector<bool> vis(n, false);
        int left = 0, right = -1;

        for (int i=0; i<n; ++i) {
            if (cnt < k) {
                pq.push({i, nums[i]});
                right++;
                cnt++;
            }
            if (cnt >= k) {
                while(vis[pq.top().first] == true) pq.pop();
                res.push_back(pq.top().second);
                vis[left++] = true;
                cnt--;
            }
        }
        return res;
    }
};

代碼二：

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        multiset<int> st;
        vector<int> res;
        int n = nums.size();
        for (int i=0; i<n; ++i) {
            if (i >= k) st.erase(st.find(nums[i-k]));
            st.insert(nums[i]);
            if (i >= k-1) res.push_back(*st.rbegin());
        }
        return res;
    }
};