【leetcode第 165 场周赛】统计全为 1 的正方形子矩阵
- 2019 年 12 月 3 日
- 筆記
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本文链接：https://blog.csdn.net/shiliang97/article/details/103334424
5277. Count Square Submatrices with All Ones
Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.
Example 1:
Input: matrix = [ [0,1,1,1], [1,1,1,1], [0,1,1,1] ] Output: 15 Explanation: There are 10 squares of side 1. There are 4 squares of side 2. There is 1 square of side 3. Total number of squares = 10 + 4 + 1 = 15. Example 2:
Input: matrix = [ [1,0,1], [1,1,0], [1,1,0] ] Output: 7 Explanation: There are 6 squares of side 1. There is 1 square of side 2. Total number of squares = 6 + 1 = 7.
Constraints:
1 <= arr.length <= 300 1 <= arr[0].length <= 300 0 <= arr[i][j] <= 1
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/count-square-submatrices-with-all-ones 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
暴力破解
选好一个起点，每个数都会当一次起点
然后从起点开始，一圈一圈往外扩展，扩展一圈就+1
直到遇到0，就结束扩展，更换下一个起点，遍历所有起点需要 N*M，然后每个起点最多查询N*M的信息，所以复杂度就是 N*M*N*M
class Solution { public: int countSquares(vector<vector<int>>& matrix) { int count=0; int x = matrix.size(); int y = matrix[0].size(); for(int a=0;a<x;a++){ for(int b=0;b<y;b++){ int flag=1; for(int t=0;b+t<y&&a+t<x;t++){ for(int l=0;l<=t;l++){ flag=flag&matrix[a+t][b+l]&&matrix[a+l][b+t]; } flag=flag&matrix[a+t][b+t]; if(flag){ //cout<<a<<b<<t<<endl; count++; }else{ break; } } } } return count; } };
