240. Search a 2D Matrix II(divide and conquer)
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1. 题目描述
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], [18, 21, 23, 26, 30]]Given target = 5, return true.Given target = 20, return false.
2. 算法分析
最糟糕的想法当然是暴力搜索了。但是,我们观察一下,每一行都是向右递增的,而每一列都是向下递增的,这说明已经为我们的解法“排好序”了。这样就很容易些想到结合有序表查找算法。采用分治的思想,将大问题逐步转化为小问题,更小的问题,舍去不必要的运算。从第一行开始找到第一个大于或等于target的数,如果等于那么完成搜索,如果大于,那么说明后面的全部大于(包括列对应的行的所有数),因此问题可以缩小为前面查找过的列对应行形成的子矩阵,采用递归的做法即可。
### 3. 算法设计
#include<iostream>#include<vector>using namespace std;bool searchMatrixRecursive(vector<vector<int> >&matrix, int row, int col, int target) { for(int r = row; r < matrix.size(); r++) { for(int c = 0; c < col; c++) { if(matrix[r][c] > target) { return searchMatrixRecursive(matrix, r+1, c, target); } else if(matrix[r][c] == target) { return true; } } } return false;}bool searchMatrix(vector<vector<int> >& matrix, int target) { if(matrix.empty()) return false; else return searchMatrixRecursive(matrix, 0, matrix[0].size(), target);}int main() { vector<vector<int> > matrix; for(int i = 0; i < 6; i++) { vector<int> tmp; for(int j = 0; j < 5; j++) { tmp.push_back((j+1)*(i+1)); } cout << endl; matrix.push_back(tmp); } int target = 8; cout << searchMatrix(matrix, target) << endl;}
**值得注意的是,不要忘记判断矩阵为空的情况!**
4. 算法优化
Accept后发现这个算法的性能并不好,如下图所示
时间复杂度和空间复杂度都很高(待优化)
注意到使用了尾递归,改造尾递归为迭代,得到更好的性能。
bool searchMatrix(vector<vector<int> >& matrix, int target) { if(matrix.empty()) { return false; } int m = matrix[0].size()-1, n = 0; while(n < matrix.size() && m >= 0) { if(matrix[n][m] == target) return true; else if( matrix[n][m] < target) n++; else m--; } return false; }
时间上改进了700多ms,可见尾递归确实浪费时间和空间开销。
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