[Swift]LeetCode598. 范围求和 II | Range Addition II

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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.

---

给定一个初始元素全部为 0，大小为 m*n 的矩阵 M 以及在 M 上的一系列更新操作。

操作用二维数组表示，其中的每个操作用一个含有两个正整数 a 和 b的数组表示，含义是将所有符合 0 <= i < a 以及 0 <= j < b 的元素 M[i][j] 的值都增加 1。

在执行给定的一系列操作后，你需要返回矩阵中含有最大整数的元素个数。

示例 1:

输入: 
m = 3, n = 3
operations = [[2,2],[3,3]]
输出: 4
解释: 
初始状态, M = 
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]

执行完操作 [2,2] 后, M = 
[[1, 1, 0],
 [1, 1, 0],
 [0, 0, 0]]

执行完操作 [3,3] 后, M = 
[[2, 2, 1],
 [2, 2, 1],
 [1, 1, 1]]

M 中最大的整数是 2, 而且 M 中有4个值为2的元素。因此返回 4。

注意:

1. m 和 n 的范围是 [1,40000]。
2. a 的范围是 [1,m]，b 的范围是 [1,n]。
3. 操作数目不超过 10000。

---

Runtime: 44 ms

Memory Usage: 19.1 MB

class Solution {
    func maxCount(_ m: Int, _ n: Int, _ ops: [[Int]]) -> Int {
        var m = m
        var n = n
        for op in ops
        {
            m = min(m, op[0])
            n = min(n, op[1])
        }
        return m * n
    }
}

---

56ms

class Solution {
    func maxCount(_ m: Int, _ n: Int, _ ops: [[Int]]) -> Int {        
        var minX = m
        var minY = n
        for operation in ops where operation[0] > 0 && operation[1] > 0 {
            minX = min(minX, operation[0])
            minY = min(minY, operation[1])
        }
        
        return minX * minY
    }
}

---

116ms

class Solution {
    func maxCount(_ m: Int, _ n: Int, _ ops: [[Int]]) -> Int {
        guard !ops.isEmpty else {
            return m * n
        }
        
        var a = Int.max
        var b = Int.max
        for op in ops {
            a = min(op.first!, a)
            b = min(op.last!, b)
        }
        
        return a * b
    }
}