leetcode 598. Range Addition II
2018-03-01 09:09
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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.
题目大意:给出矩阵大小,给出每次加的坐标(x,y)那么每次操作都会使(0~x,0~y)这个矩阵内的格子加1,求最大的格子的个数。
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: The range of m and n is [1,40000]. The range of a is [1,m], and the range of b is [1,n]. The range of operations size won't exceed 10,000.
题目大意:给出矩阵大小,给出每次加的坐标(x,y)那么每次操作都会使(0~x,0~y)这个矩阵内的格子加1,求最大的格子的个数。
class Solution { public: int maxCount(int m, int n, vector<vector<int>>& ops) { vector<pair<int, int> > v; if (ops.size() == 0) return m*n; int x = 100000; int y = 100000; for (int i = 0; i < ops.size(); ++i) { x = min(x, ops[i][0]); y = min(y, ops[i][1]); } return x*y; } };
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