[LeetCode] Range Addition II 范围相加之二
2017-06-09 23:42
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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.
Example 1:
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.
这道题看起来像是之前那道Range Addition的拓展,但是感觉实际上更简单一些。每次在ops中给定我们一个横纵坐标,将这个子矩形范围内的数字全部自增1,让我们求最大数字的个数。原数组初始化均为0,那么如果ops为空,没有任何操作,那么直接返回m*n即可,我们可以用一个优先队列来保存最大数字矩阵的横纵坐标,我们可以通过举些例子发现,只有最小数字组成的边界中的数字才会被每次更新,所以我们想让最小的数字到队首,更优先队列的排序机制是大的数字在队首,所以我们对其取相反数,这样我们最后取出两个队列的队首数字相乘即为结果,参见代码如下:
解法一:
我们可以对空间进行优化,不使用优先队列,而是每次用ops中的值来更新m和n,取其中较小值,这样遍历完成后,m和n就是最大数矩阵的边界了,参见代码如下:
解法二:
类似题目:
Range Addition
参考资料:
https://discuss.leetcode.com/topic/90540/c-java-clean-code
LeetCode All in One 题目讲解汇总(持续更新中...)
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.
这道题看起来像是之前那道Range Addition的拓展,但是感觉实际上更简单一些。每次在ops中给定我们一个横纵坐标,将这个子矩形范围内的数字全部自增1,让我们求最大数字的个数。原数组初始化均为0,那么如果ops为空,没有任何操作,那么直接返回m*n即可,我们可以用一个优先队列来保存最大数字矩阵的横纵坐标,我们可以通过举些例子发现,只有最小数字组成的边界中的数字才会被每次更新,所以我们想让最小的数字到队首,更优先队列的排序机制是大的数字在队首,所以我们对其取相反数,这样我们最后取出两个队列的队首数字相乘即为结果,参见代码如下:
解法一:
class Solution { public: int maxCount(int m, int n, vector<vector<int>>& ops) { if (ops.empty() || ops[0].empty()) return m * n; priority_queue<int> r, c; for (auto op : ops) { r.push(-op[0]); c.push(-op[1]); } return r.top() * c.top(); } };
我们可以对空间进行优化,不使用优先队列,而是每次用ops中的值来更新m和n,取其中较小值,这样遍历完成后,m和n就是最大数矩阵的边界了,参见代码如下:
解法二:
class Solution { public: int maxCount(int m, int n, vector<vector<int>>& ops) { for (auto op : ops) { m = min(m, op[0]); n = min(n, op[1]); } return m * n; } };
类似题目:
Range Addition
参考资料:
https://discuss.leetcode.com/topic/90540/c-java-clean-code
LeetCode All in One 题目讲解汇总(持续更新中...)
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