题目描述：

LeetCode 598. Range Addition II

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.

题目大意：

给定m * n矩阵M，初始为0，然后执行一些更新操作。

数组ops表示一组更新操作，每一个操作(a, b)，表示将矩阵0 <= i < a 并且 0 <= j < b的区域值+1。

进行若干操作后，求矩阵的最大值。

注意：

1. m和n的范围[1, 40000]
2. a的范围[1, m]，b的范围[1, n]
3. 操作不超过10000个

解题思路：

求ops[0 .. len][0]和ops[0 .. len][1]的最小值

Python代码：

class Solution(object):
    def maxCount(self, m, n, ops):
        """
        :type m: int
        :type n: int
        :type ops: List[List[int]]
        :rtype: int
        """
        if not ops: return m * n
        return min(op[0] for op in ops) * min(op[1] for op in ops)

 

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