LeetCode 598. Range Addition II

Given an m * n matrixMinitialized with all0's and several update operations.

Operations are represented by a 2D array, and each operation is represented by an array with twopositiveintegersaandb, which meansM[i][j]should beadded by onefor all0 <= i < aand0 <= 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.
   

题意：给定一个m * n的矩阵M，首先将矩阵M的所有元素初始化为0，然后对矩阵M进行更新操作。
更新操作的定义：每个操作用一个由两个正整数a、b组成的数组[a,b]表示，i表示矩阵M的行，j表示矩阵M的列;
[a,b]的含义为：对1 <= i <= a，1 <= j <= b的所有元素进行加1操作。
执行完所有操作后，返回矩阵中最大整数的个数。

思路：即相当于求二维数组ops中每一列的最小值

public int maxCount(int m, int n, int[][] ops) {
        if(ops == null)
            return 0;
        int row = Integer.MAX_VALUE, col = Integer.MAX_VALUE;
        for(int i = 0; i < ops.length; i++){
            if(row > ops[i][0])
                row = ops[i][0];
            if(col > ops[i][1])
                col = ops[i][1];
        }
        return row * col;
    }

LeetCode提供的方法思路一样，但算法更简洁

public int maxCount(int m, int n, int[][] ops){
        for(int[] op : ops){
            m = Math.min(m, op[0]);
            n = Math.min(n, op[1]);
        }
        return m * n;
    }