《Cracking the Coding Interview》——第18章：难题——题目11

2014-04-29 04:30

题目：给定一个由‘0’或者‘1’构成的二维数组，找出一个四条边全部由‘1’构成的正方形（矩形中间可以有‘0’），使得矩形面积最大。

解法：用动态规划思想，记录二维数组每个元素向上下左右四个方向各有多少个连续的‘1’，然后用O(n^3)时间计算出满足条件的最大正方形。时间复杂度O(n^3)，空间复杂度O(n^2)。

代码：

// 18.11 Given an NxN matrix of 0s and 1s, find out a subsquare whose all four borders are all 1s. If multiple satisfies the condition, any one is OK.
// I'll return the size and the left top corner of the subsquare.
#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    void maxSubsquare(const vector<vector<int> > &matrix, int &max_left, int &max_top, int &max_size) {
        int n = matrix.size();
        
        max_left = max_top = max_size = -1;
        
        if (n <= 1) {
            return;
        }
        
        vector<vector<int> > top   (n, vector<int>(n));
        vector<vector<int> > bottom(n, vector<int>(n));
        vector<vector<int> > left  (n, vector<int>(n));
        vector<vector<int> > right (n, vector<int>(n));
        
        int i, j;
        int tmp;
        
        // use DP to preprocess the data, count how many consecutive 1s are there to the left, right, top, bottom of matrix[i][j].
        for (i = 0; i <= n - 1; ++i) {
            tmp = 0;
            for (j = 0; j <= n - 1; ++j) {
                left[i][j] = matrix[i][j] ? (++tmp) : (tmp = 0);
            }
        }
        for (j = 0; j <= n - 1; ++j) {
            tmp = 0;
            for (i = 0; i <= n - 1; ++i) {
                top[i][j] = matrix[i][j] ? (++tmp) : (tmp = 0);
            }
        }
        for (i = n - 1; i >= 0; --i) {
            tmp = 0;
            for (j = n - 1; j >= 0; --j) {
                right[i][j] = matrix[i][j] ? (++tmp) : (tmp = 0);
            }
        }
        for (j = n - 1; j >= 0; --j) {
            tmp = 0;
            for (i = n - 1; i >= 0; --i) {
                bottom[i][j] = matrix[i][j] ? (++tmp) : (tmp = 0);
            }
        }
        
        int len;
        // O(n ^ 3) solution with O(n ^ 2) space usage.
        for (i = 0; i < n; ++i) {
            for (j = 0; j < n; ++j) {
                for (len = 2; len + i <= n && len + j <= n; ++len) {
                    if (right[i][j] < len || bottom[i][j] < len) {
                        continue;
                    }
                    if (left[i][j + len - 1] < len || bottom[i][j + len - 1] < len) {
                        continue;
                    }
                    if (right[i + len - 1][j] < len || top[i + len - 1][j] < len) {
                        continue;
                    }
                    if (left[i + len - 1][j + len - 1] < len || top[i + len - 1][j + len - 1] < len) {
                        continue;
                    }
                    // all four borders are '1's.
                    if (len > max_size) {
                        max_top = i;
                        max_left = j;
                        max_size = len;
                    }
                }
            }
        }
        
        // clear up data
        for (i = 0; i < n; ++i) {
            left[i].clear();
            right[i].clear();
            top[i].clear();
            bottom[i].clear();
        }
        left.clear();
        right.clear();
        top.clear();
        bottom.clear();
    };
};

int main()
{
    int n;
    int i, j;
    vector<vector<int> > matrix;
    Solution sol;
    int max_left, max_top, max_size;
    
    while (cin >> n && n > 0) {
        matrix.resize(n);
        for (i = 0; i < n; ++i) {
            matrix[i].resize(n);
        }
        
        for (i = 0; i < n; ++i) {
            for (j = 0; j < n; ++j) {
                cin >> matrix[i][j];
            }
        }
        
        sol.maxSubsquare(matrix, max_left, max_top, max_size);
        if (max_size > 0) {
            cout << max_top << ' ' << max_left << endl;
            cout << max_top << ' ' << max_left + max_size - 1 << endl;
            cout << max_top + max_size - 1 << ' ' << max_left << endl;
            cout << max_top + max_size - 1 << ' ' << max_left + max_size - 1 << endl;
        } else {
            cout << "No subsquare found." << endl;
        }
        
        for (i = 0; i < n; ++i) {
            matrix[i].clear();
        }
        matrix.clear();
    }
    
    return 0;
}