[Leetcode] Search a 2D Matrix

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 from left to right.
• The first integer of each row is greater than the last integer of the previous row.

 

For example,

Consider the following matrix:

[

  [1,   3,  5,  7],

  [10, 11, 16, 20],

  [23, 30, 34, 50]

]

Given target = 3, return true.

 

从右上角向左下角搜索，初始row = 0, col = n - 1; 如果当前值等于目标值，返回真，如果以目标值大，则舍弃当前列，否则舍弃当前行。

 1 class Solution {

 2 public:

 3     bool searchMatrix(vector<vector<int> > &matrix, int target) {

 4         int row = 0, col = matrix[0].size() - 1;

 5         while (row < matrix.size() && col >=0) {

 6             if (matrix[row][col] == target) {

 7                 return true;

 8             } else if (matrix[row][col] > target) {

 9                 --col;

10             } else {

11                 ++row;

12             }

13         }

14         return false;

15     }

16 };

 第二遍做的时候发现其实这个矩阵不完全是杨氏矩阵，可以使用二分搜索，下面是代码。

 1 class Solution {

 2 public:

 3     bool searchMatrix(vector<vector<int> > &matrix, int target) {

 4         int m = matrix.size();

 5         int n = matrix[0].size();

 6         int L = 0, R = n * m - 1, M;

 7         int col, row;

 8         while (L <= R) {

 9             M = L + ((R - L) >> 1);

10             row = M / n;

11             col = M % n;

12             if (matrix[row][col] > target) R = M - 1;

13             else if (matrix[row][col] < target) L = M + 1;

14             else return true;

15         }

16         return false;

17     }

18 };

 上面是对所有元素二分，复杂度是log(n*m)，还可以更优化一点，对于普通的杨氏矩阵也适用，复杂度为log(n)+log(m).

 1 class Solution {

 2 public:

 3     bool searchMatrix(vector<vector<int> > &matrix, int target) {

 4         int row = lowerBound(matrix, target);

 5         if (row == -1) return false;

 6         return binSearch(matrix, target, row);

 7     }

 8     

 9     int lowerBound(vector<vector<int> > &matrix, int target) {

10         int L = 0, R = matrix.size() - 1, M;

11         while (L <= R) {

12             M = L + ((R - L) >> 1);

13             if (matrix[M][0] <= target) L = M + 1;

14             else R = M - 1;

15         }

16         return R;

17     }

18     

19     bool binSearch(vector<vector<int> > &matrix, int target, int row) {

20         int L = 0, R = matrix[row].size() - 1, M;

21         while (L <= R) {

22             M = L + ((R - L) >> 1);

23             if (matrix[row][M] < target) L = M + 1;

24             else if (matrix[row][M] > target) R = M - 1;

25             else return true;

26         }

27         return false;

28     }

29 };