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 = 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.

public class Solution {    public int maxCount(int m, int n, int[][] ops) {        if(ops==null||ops.length==0)return m*n;        int minx=ops[0][0],miny=ops[0][1];        for(int i=1;i<ops.length;i++)        {            minx=Math.min(ops[i][0],minx);            miny=Math.min(ops[i][1],miny);        }        return minx*miny;    }}

取区域内行列最小值就可以圈定范围