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 }
《Cracking the Coding Interview》——第18章：难题——题目12,布布扣,bubuko.com
时间： 2024-12-25 20:55:05