[抄题]:
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.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
[一句话思路]:
行列都取最小值
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O() Space complexity: O()
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
表示每次取出的是数组,op[0]表示该被取出的数组中的第0位,op[1]表示该数组中的第1位
//find min, max for (int[] op : ops) { row = Math.min(op[0], row); col = Math.min(op[1], col); }
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
370. Range Addition 就是操作数组吧
[代码风格] :
class Solution { public int maxCount(int m, int n, int[][] ops) { //cc if (ops == null || ops.length == 0) return m * n; //ini:min max int row = Integer.MAX_VALUE, col = Integer.MAX_VALUE; //find min, max for (int[] op : ops) { row = Math.min(op[0], row); col = Math.min(op[1], col); } return row * col; } }