LeetCode Range Addition II

原题链接在这里：https://leetcode.com/problems/range-addition-ii/description/

题目：

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.

题解：

找出最小的更新row, 和最小的更新column, 返回乘机. 记得这两个可能比对应的m, n大, 所以初始值时m,n.

Time Complexity: O(ops.length). Space: O(1).

AC Java:

 class Solution {
     public int maxCount(int m, int n, int[][] ops) {
         if(ops == null || ops.length == 0){
             return m*n;
         }

         int minRow = m;
         int minColumn = n;
         for(int [] op : ops){
             minRow = Math.min(minRow, op[0]);
             minColumn = Math.min(minColumn, op[1]);
         }
         return minRow*minColumn;
     }
 }

类似Range Addition.