LeetCode - 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 = 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.
模拟会空间复杂度会超,必须直接计算。计算重叠部分。
class Solution { public int maxCount(int m, int n, int[][] ops) { if (ops == null || ops.length <=0 || ops[0].length <= 0) return m*n; int minX = Integer.MAX_VALUE, minY = Integer.MAX_VALUE; for (int i=0; i<ops.length; i++) { int p = ops[i][0], q = ops[i][1]; if (minX > p) minX = p; if (minY > q) minY = q; } return minX * minY; } }
作者:Pickle
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