LeetCode – 598. Range Addition II

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.
 这道题看上去很唬人,不过纸上比划一下发现其实非常简单 —— 就是求最小的范围(注意是范围,我一开始简单的想成求最小的区块)

public class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        if(ops.length == 0) return m*n;
        int r = ops.length-1;
        int minrow = Integer.MAX_VALUE;
        int mincol = Integer.MAX_VALUE;
        while(r>=0){
            if(ops[r][0] < minrow){
                minrow = ops[r][0];
            }
            if(ops[r][1] < mincol){
                mincol = ops[r][1];
            }
            r--;
        }
        return mincol * minrow;
    }
}

喜欢的话订阅一个呗~第一时间收到文章更新哟~

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