598 Range Addition II 范圍求和 II
Description:
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:
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.
題目描述:
給定一個初始元素全部為 0疲牵，大小為 m*n 的矩陣 M 以及在 M 上的一系列更新操作笤喳。
操作用二維數(shù)組表示刊咳，其中的每個操作用一個含有兩個正整數(shù) a 和 b 的數(shù)組表示及志，含義是將所有符合 0 <= i < a 以及 0 <= j < b 的元素 M[i][j] 的值都增加 1修噪。
在執(zhí)行給定的一系列操作后抡秆，你需要返回矩陣中含有最大整數(shù)的元素個數(shù)着降。
示例 :
示例 1:
輸入:
m = 3, n = 3
operations = [[2,2],[3,3]]
輸出: 4
解釋:
初始狀態(tài), M =
[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
執(zhí)行完操作 [2,2] 后, M =
[[1, 1, 0],
[1, 1, 0],
[0, 0, 0]]
執(zhí)行完操作 [3,3] 后, M =
[[2, 2, 1],
[2, 2, 1],
[1, 1, 1]]
M 中最大的整數(shù)是 2, 而且 M 中有4個值為2的元素化漆。因此返回 4。
注意:
m 和 n 的范圍是 [1,40000]昙啄。
a 的范圍是 [1,m]穆役，b 的范圍是 [1,n]。
操作數(shù)目不超過 10000梳凛。
思路:
取到所有長度和寬度的最小值即可
時(shí)間復(fù)雜度O(n), 空間復(fù)雜度O(1)
代碼:
C++:
class Solution
{
public:
int maxCount(int m, int n, vector<vector<int>>& ops)
{
for (auto v : ops)
{
m = min(m, v[0]);
n = min(n, v[1]);
}
return m * n;
}
};
Java:
class Solution {
public int maxCount(int m, int n, int[][] ops) {
for (int[] v : ops) {
m = Math.min(m, v[0]);
n = Math.min(n, v[1]);
}
return m * n;
}
}
Python:
class Solution:
def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
return min((i[0] for i in ops)) * min((i[1] for i in ops)) if ops else m * n
