LeetCode刷题之Range Addition II

Problem

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:

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

My Solution

class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        int a = -1, b = -1;
        int rMin = Integer.MAX_VALUE, cMin = Integer.MAX_VALUE;
        if (ops.length == 0) {
            return m * n;
        }
        for (int oi = 0; oi < ops.length; ++oi) {
            a = ops[oi][0];
            b = ops[oi][1];
            if (a < rMin) {
                rMin = a;
            }
            if (b < cMin) {
                cMin = b;
            }
        }
        return rMin * cMin;
    }
}

Great Solution

public class Solution {
    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;
    }
}