《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 == ) {
             return ;
         }
         m = matrix[].size();
         if (m == ) {
             return ;
         }

         int i, j, k;
         vector<int> v;
         int msum;
         int sum;

         v.resize(m);
         msum = INT_MIN;
         for (i = ; i < n; ++i) {
             fill(v.begin(), v.end(), );
             for (j = i; j < n; ++j) {
                 for (k = ; 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 = ; i < n; ++i) {
             if (v[i] >= ) {
                 msum = max(msum, v[i]);
                 break;
             }
         }
         if (i == n) {
             return msum;
         }

         msum = sum = ;
         for (i = ; i < n; ++i) {
             sum += v[i];
             msum = max(msum, sum);
             sum = max(sum, );
         }

         return msum;
     };
 };

 int main()
 {
     int i, j;
     int n, m;
     vector<vector<int> > matrix;
     Solution sol;

     while (cin >> n >> m && (n >  && m > )) {
         matrix.resize(n);
         for (i = ; i < n; ++i) {
             matrix[i].resize(m);
             for (j = ; j < m; ++j) {
                 cin >> matrix[i][j];
             }
         }
         cout << sol.largestSubmatrixSum(matrix) << endl;

         for (i = ; i < n; ++i) {
             matrix[i].clear();
         }
         matrix.clear();
     }

     return ;
 }