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.
题目标签：Math
这道题目给了我们一个 m*n 的matrix， 起初都是0， 根据operation给其中一部分区域加1。最后要return 最大值integer的个数。
我们可以从另一个角度出发，把这个题目转化成图形来理解，最大的值的区域就是所有operation的交集。如何找到这个区域呢，我们需要记录一个min x 和min y 来求出交集的区域 = x*y， 相当于在求面积。
举两个例子来看一下：
Example 1:
maxCount(3,3,[ [1,2], [2,1] ])
0 0 0 1 1 0 2 1 0
0 0 0 [1,2] -> 0 0 0 [2,1] -> 1 0 0 return 1;
0 0 0 0 0 0 0 0 0
最小的 x = 1， 最小的 y = 1, 所以最小的交集是 0,0 这个坐标， 它的区域 = 1 * 1。
Example 2:
maxCount(3,3,[ [1,3], [2,2] ])
0 0 0 1 1 1 2 2 1
0 0 0 [1,3] -> 0 0 0 [2,2] -> 1 1 0 return 2;
0 0 0 0 0 0 0 0 0
最小的 x = 1， 最小的 y = 2, 所以最小的交集是 0,0 和 0,1 这两个坐标， 它的区域 = 1 * 2。
Java Solution:
Runtime beats 77.83%
完成日期：06/17/2017
关键词：matrix, 2d array
关键点：把题目转换成图形帮助理解
1 public class Solution
2 {
3 public int maxCount(int m, int n, int[][] ops)
4 {
5 if(ops == null || ops.length == 0 )
6 return m*n;
7
8 int x = Integer.MAX_VALUE;
9 int y = x;
10
11
12
13 for(int i=0; i<ops.length; i++)
14 {
15 if(ops[i][0] < x)
16 x = ops[i][0];
17 if(ops[i][1] < y)
18 y = ops[i][1];
19
20 }
21
22 return x*y;
23 }
24 }
参考资料：
N / A