《Cracking the Coding Interview》——第18章：难题——题目12

2014-04-29 04:36

题目：最大子数组和的二位扩展：最大子矩阵和。

解法：一个维度上进行枚举，复杂度O(n^2)；另一个维度执行最大子数组和算法，复杂度O(n)。总体时间复杂度为O(n^3)，还需要O(n)额外空间。

代码：

// 18.12 Given an n x n matrix, find the submatrix with largest sum. Return the sum as the result.
#include <algorithm>
#include <climits>
#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int largestSubmatrixSum (const vector<vector<int> > &matrix) {
        n = matrix.size();
        if (n == 0) {
            return 0;
        }
        m = matrix[0].size();
        if (m == 0) {
            return 0;
        }
        
        int i, j, k;
        vector<int> v;
        int msum;
        int sum;
        
        v.resize(m);
        msum = INT_MIN;
        for (i = 0; i < n; ++i) {
            fill(v.begin(), v.end(), 0);
            for (j = i; j < n; ++j) {
                for (k = 0; k < m; ++k) {
                    v[k] += matrix[j][k];
                }
                sum = maxSubarraySum(v, m);
                msum = max(msum, sum);
            }
        }
        v.clear();
        return msum;
    };
private:
    int n, m;
    
    int maxSubarraySum(const vector<int> &v, int n) {
        int msum;
        int sum;
        int i;
        
        msum = INT_MIN;
        for (i = 0; i < n; ++i) {
            if (v[i] >= 0) {
                msum = max(msum, v[i]);
                break;
            }
        }
        if (i == n) {
            return msum;
        }
        
        msum = sum = 0;
        for (i = 0; i < n; ++i) {
            sum += v[i];
            msum = max(msum, sum);
            sum = max(sum, 0);
        }
        
        return msum;
    };
};

int main()
{
    int i, j;
    int n, m;
    vector<vector<int> > matrix;
    Solution sol;
    
    while (cin >> n >> m && (n > 0 && m > 0)) {
        matrix.resize(n);
        for (i = 0; i < n; ++i) {
            matrix[i].resize(m);
            for (j = 0; j < m; ++j) {
                cin >> matrix[i][j];
            }
        }
        cout << sol.largestSubmatrixSum(matrix) << endl;
        
        for (i = 0; i < n; ++i) {
            matrix[i].clear();
        }
        matrix.clear();
    }
    
    return 0;
}