算法分析与设计——LeetCode Problem.598 Range Addition II

问题详情

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.

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问题分析及思路

题意比较简单，一开始我以为可以通过遍历，三层循环，每次每个数加1。然后再将所有的数与在[0][0]的数比较，得到最大的计数。
后来发现固定的是[0][0]的数最大，因此只需要求出在所有的operations 数组中对应的最小值，即可得到最小并的行与列数。

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具体代码

class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        int min_x = m;
        int min_y = n;
        int i;
        for(i = 0; i < ops.size(); i++) {
            if(min_x > ops.at(i).at(0)) min_x=ops.at(i).at(0);
            if(min_y > ops.at(i).at(1)) min_y=ops.at(i).at(1);
        }
        return min_x * min_y;
    }
};