[LeetCode] 598. Range Addition II 范圍相加之二

 

Given an m * n matrix M initialized with all 0's and several update operations.

Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.

You need to count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:

Input: 
m = 3, n = 3
operations = [[2,2],[3,3]]
Output: 4
Explanation: 
Initially, M = 
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]

After performing [2,2], M = 
[[1, 1, 0],
 [1, 1, 0],
 [0, 0, 0]]

After performing [3,3], M = 
[[2, 2, 1],
 [2, 2, 1],
 [1, 1, 1]]

So the maximum integer in M is 2, and there are four of it in M. So return 4.

 

Note:

1. The range of m and n is [1,40000].
2. The range of a is [1,m], and the range of b is [1,n].
3. The range of operations size won't exceed 10,000.

 

這道題看起來像是之前那道 Range Addition 的拓展，但是感覺實際上更簡單一些。每次在 ops 中給定我們一個橫縱坐標，將這個子矩形范圍內的數字全部自增1，讓我們求最大數字的個數。原數組初始化均為0，那么如果 ops 為空，沒有任何操作，那么直接返回 m*n 即可，我們可以用一個優先隊列來保存最大數字矩陣的橫縱坐標，我們可以通過舉些例子發現，只有最小數字組成的邊界中的數字才會被每次更新，所以我們想讓最小的數字到隊首，更優先隊列的排序機制是大的數字在隊首，所以我們對其取相反數，這樣我們最后取出兩個隊列的隊首數字相乘即為結果，參見代碼如下：

 

解法一：

class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        if (ops.empty() || ops[0].empty()) return m * n;
        priority_queue<int> r, c;
        for (auto op : ops) {
            r.push(-op[0]);
            c.push(-op[1]);
        }
        return r.top() * c.top();
    }
};

 

我們可以對空間進行優化，不使用優先隊列，而是每次用 ops 中的值來更新m和n，取其中較小值，這樣遍歷完成后，m和n就是最大數矩陣的邊界了，參見代碼如下：

 

解法二：

class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        for (auto op : ops) {
            m = min(m, op[0]);
            n = min(n, op[1]);
        }
        return m * n;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/598

 

類似題目：

Range Addition

 

參考資料：

https://leetcode.com/problems/range-addition-ii/

https://leetcode.com/problems/range-addition-ii/discuss/103595/Java-Solution-find-Min

 

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