mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2026-03-13 10:02:05 +08:00
82 lines
2.7 KiB
Markdown
82 lines
2.7 KiB
Markdown
# [598. Range Addition II](https://leetcode.com/problems/range-addition-ii/)
|
||
|
||
|
||
## 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.
|
||
|
||
**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。
|
||
|
||
注意:
|
||
|
||
- m 和 n 的范围是 [1,40000]。
|
||
- a 的范围是 [1,m]，b 的范围是 [1,n]。
|
||
- 操作数目不超过 10000。
|
||
|
||
|
||
## 解题思路
|
||
|
||
- 给定一个初始都为 0 的 m * n 的矩阵，和一个操作数组。经过一系列的操作以后，最终输出矩阵中最大整数的元素个数。每次操作都使得一个矩形内的元素都 + 1 。
|
||
- 这一题乍一看像线段树的区间覆盖问题，但是实际上很简单。如果此题是任意的矩阵，那就可能用到线段树了。这一题每个矩阵的起点都包含 [0 , 0] 这个元素，也就是说每次操作都会影响第一个元素。那么这道题就很简单了。经过 n 次操作以后，被覆盖次数最多的矩形区间，一定就是最大整数所在的区间。由于起点都是第一个元素，所以我们只用关心矩形的右下角那个坐标。右下角怎么计算呢？只用每次动态的维护一下矩阵长和宽的最小值即可。
|
||
|
||
## 代码
|
||
|
||
```go
|
||
|
||
package leetcode
|
||
|
||
func maxCount(m int, n int, ops [][]int) int {
|
||
minM, minN := m, n
|
||
for _, op := range ops {
|
||
minM = min(minM, op[0])
|
||
minN = min(minN, op[1])
|
||
}
|
||
return minM * minN
|
||
}
|
||
|
||
func min(a, b int) int {
|
||
if a < b {
|
||
return a
|
||
}
|
||
return b
|
||
}
|
||
|
||
``` |