《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：

 1 // 18.12 Given an n x n matrix, find the submatrix with largest sum. Return the sum as the result.

 2 #include <algorithm>

 3 #include <climits>

 4 #include <iostream>

 5 #include <vector>

 6 using namespace std;

 7 

 8 class Solution {

 9 public:

10     int largestSubmatrixSum (const vector<vector<int> > &matrix) {

11         n = matrix.size();

12         if (n == 0) {

13             return 0;

14         }

15         m = matrix[0].size();

16         if (m == 0) {

17             return 0;

18         }

19         

20         int i, j, k;

21         vector<int> v;

22         int msum;

23         int sum;

24         

25         v.resize(m);

26         msum = INT_MIN;

27         for (i = 0; i < n; ++i) {

28             fill(v.begin(), v.end(), 0);

29             for (j = i; j < n; ++j) {

30                 for (k = 0; k < m; ++k) {

31                     v[k] += matrix[j][k];

32                 }

33                 sum = maxSubarraySum(v, m);

34                 msum = max(msum, sum);

35             }

36         }

37         v.clear();

38         return msum;

39     };

40 private:

41     int n, m;

42     

43     int maxSubarraySum(const vector<int> &v, int n) {

44         int msum;

45         int sum;

46         int i;

47         

48         msum = INT_MIN;

49         for (i = 0; i < n; ++i) {

50             if (v[i] >= 0) {

51                 msum = max(msum, v[i]);

52                 break;

53             }

54         }

55         if (i == n) {

56             return msum;

57         }

58         

59         msum = sum = 0;

60         for (i = 0; i < n; ++i) {

61             sum += v[i];

62             msum = max(msum, sum);

63             sum = max(sum, 0);

64         }

65         

66         return msum;

67     };

68 };

69 

70 int main()

71 {

72     int i, j;

73     int n, m;

74     vector<vector<int> > matrix;

75     Solution sol;

76     

77     while (cin >> n >> m && (n > 0 && m > 0)) {

78         matrix.resize(n);

79         for (i = 0; i < n; ++i) {

80             matrix[i].resize(m);

81             for (j = 0; j < m; ++j) {

82                 cin >> matrix[i][j];

83             }

84         }

85         cout << sol.largestSubmatrixSum(matrix) << endl;

86         

87         for (i = 0; i < n; ++i) {

88             matrix[i].clear();

89         }

90         matrix.clear();

91     }

92     

93     return 0;

94 }