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 = 3operations = [[2,2],[3,3]]Output: 4Explanation: 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.

题意：

这道题的题意比较难理解，意思是给你一个m*n的全0数组，给你一组数组ops，ops里面的每一组数组，都要在m*n的数组上画一个对应的框，框内的数自增一，直到ops里面的数组都加完，然后输出最大值的个数

算法思路：

题意可以理解为给了一张M*N的白纸，在上面不断地画框，最后框最多的就是结果

所以只需要对每行每列的数取最小值，最终的最小值相乘就是个二维里面的最小值

代码：

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;    }}