【数学】C033_范围求和II（暴力 | 脑筋急转弯）

一、题目描述

Given an mn 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.

二、题解

(1) 暴力（超时）

/**
 * @thought：暴力破解（超时）：
 *
 * 因为a>0，b>0，故每一个操作都会涉及到M[0][0]，故我们完成每个操作后，
 * 再次遍历矩阵，看有多少个元素是等于M[0][0]就知道最大元素有多少个了
 */
public int maxCount(int m, int n, int[][] ops) {
  int[][] M = new int[m][n];
  int maxCount = 0;

  for(int[] op : ops)
  for (int i = 0; i < op[0]; i++)
  for (int j = 0; j < op[1]; j++)
    M[i][j]++;

  for (int i = 0; i < m; i++)
  for (int j = 0; j < n; j++)
  if(M[i][j]==M[0][0])
    maxCount++;
  return maxCount;
}

复杂度分析

• 时间复杂度： $O(x*m*n)$， $x$ 是 $ops$ 数组的行数。
• 空间复杂度： $O(m*n)$

(2) 脑筋急转弯

/**
 * @thought：脑筋急转弯：因为a影响矩阵M的每一行的元素，b影响每一列的元素。
 * 因此，对于最小的a1和b1，其他ai和bi的操作范围都会覆盖a1,b1，
 * 故我们只需找出被操作的最小面积即可（即 a1*b1）
 *
 * @date: 1/17/2020 3:49 PM
 * @Execution info：
 *  ·执行用时 0 ms 击败了 100% 的java用户
 *  ·内存消耗 MB 击败了 % 的java用户
 * @Asymptotic Time Complexity：O()
 */
public int maxCount2(int m, int n, int[][] ops) {
  for (int[] op : ops) {
    if(op[0] < m)  m=op[0];
    if(op[1] < n)  n=op[1];
  }
  return m * n;
}

复杂度分析

• 时间复杂度： $O(x)$， $x$ 是 $ops$ 数组的行数。
• 空间复杂度： $O(1)$。