[leetcode]: 598. Range Addition II

1.题目

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.
给一个m*n的全0矩阵和一些更新操作。每个操作用两个数字a,b组成的二维数组来表示，操作定义为对矩阵所有下标满足0 <= i < a 和 0 <= j < b的元素+1。
求经过一系列操作后，矩阵中最大的元素有多少个。
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.

2.分析

求行和列下标的最小值即可。因为+1操作的范围是0 <= i < a 和 0 <= j < b，所以求交集的话找最小值即可。

3.代码

def maxCount(self, m, n, ops):    if not ops:        return 0    x=[op[0] for op in ops]    y=[op[1] for op in ops]    minx=min(x)    miny=min(y)    if minx>m:        minx=m    if miny>n:        miny=n    return minx*miny