[leetcode-598-Range Addition II]

最新推荐文章于 2025-02-22 07:15:00 发布

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最新推荐文章于 2025-02-22 07:15:00 发布

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原文链接：http://www.cnblogs.com/hellowooorld/p/6915320.html

本文介绍了一个算法问题，即给定一个初始为0的m*n矩阵，通过一系列更新操作后，如何找出矩阵中最大整数的数量。更新操作由一个二维数组表示，每个操作包括两个整数a和b，表示所有0<=i<a且0<=j<b的M[i][j]应该增加1。文章提供了一种解决方案，通过记录每次操作覆盖的最小行和列，最终返回这些操作共同覆盖的矩阵区域大小。

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.

思路：

思考可以发现，矩阵越靠近左上角的元素值越大，因为要加1的元素行和列索引是从0开始的。

那么只需要找到操作次数最多的元素位置即可。而操作次数最多的元素肯定是偏向于靠近矩阵左上角的。

int maxCount(int m, int n, vector<vector<int> >& ops) 
{
        int minrow = 50000,mincol =50000;
    for(int i =0;i<ops.size();i++)
    {
      minrow =min(minrow,ops[i][0]);
      mincol = min(mincol,ops[i][1]);
    }
    minrow = min(minrow,m);
    mincol = min(mincol,n);
    return minrow*mincol;
}