598. Range Addition II

598. Range Addition II

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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.

題意：

這道題的題意比較難理解，意思是給你一個m*n的全0數組，給你一組數組ops，ops裏面的每一組數組，都要在m*n的數組上畫一個對應的框，框內的數自增一，直到ops裏面的數組都加完，然後輸出最大值的個數

算法思路：

題意可以理解爲給了一張M*N的白紙，在上面不斷地畫框，最後框最多的就是結果

所以只需要對每行每列的數取最小值，最終的最小值相乘就是個二維裏面的最小值

代碼：

package com.bjsxt.hibernate;

import com.sun.org.apache.regexp.internal.recompile;

public class RangeAdditionII {
	
	public int maxCount(int m, int n, int[][] ops) {
        for(int[] k : ops){
        	m = Math.min(m, k[0]);
        	n = Math.min(n, k[1]);
        }
        
        return m*n;
    }
}