LintCode 853. Number Of Corner Rectangles

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于 2019-06-02 14:36:13 首次发布

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本文介绍了一种算法，用于解决在一个仅包含0和1的网格中寻找角矩形的数量的问题。通过比较每行之间的1的位置来统计矩形数量，详细解释了实现过程和代码示例。

Number Of Corner Rectangles
中文English
Given a grid with only 0 and 1, find the number of corner rectangles.

Note that only the corners need to have the value 1. Also, all four 1s used must be distinct.

Example
Example 1:

Input:
[
[1, 0, 0, 1, 0],
[0, 0, 1, 0, 1],
[0, 0, 0, 1, 0],
[1, 0, 1, 0, 1]
]
Output: 1
Explanation: There is only one corner rectangle, with corners grid[1][2], grid[1][4], grid[3][2], grid[3][4].
Example 2:

Input:
[
[1, 1, 1],
[1, 1, 1],
[1, 1, 1]
]
Output: 9
Explanation: There are four 2x2 rectangles, four 2x3 and 3x2 rectangles, and one 3x3 rectangle.
Example 3:

Input: [[1,1,1,1]]
Output: 0
Explanation: Rectangles must have four distinct corners.
Notice
The number of rows and columns of grid will each be in the range [1, 200][1,200].
Each grid[i][j] will be either 0 or 1.
The number of 1s in the grid will be at most 6000.

解法1：
思路：先看第1行，然后第2到M行逐行跟第1行比，统计有1重合的数目。假设第i行跟第1行有1重合的数目为x，则新添矩形数目为C(x,2)=x*(x-1)/2，因为这x个1两两可以对应一个矩形。然后再看第2行，同样处理，直到一直处理完第M-1行。
代码如下。

class Solution {
public:
    /**
     * @param grid: the grid
     * @return: the number of corner rectangles
     */
    int countCornerRectangles(vector<vector<int>> &grid) {
        int m = grid.size();
        if (m == 0) return 0;
        int n = grid[0].size();
        int cnt = 0, sum = 0;
        for (int i = 0; i < m - 1; i++) {
            for (int j = i + 1; j < m; j++) {
                int cnt = 0;
                for (int k = 0; k < n; k++) {
                    if (grid[i][k] == 1 && grid[j][k] == 1) {
                        cnt++;
                    }
                }
                sum += cnt * (cnt - 1) / 2;
            }
        }
        return sum;
    }
};