28. Search a 2D Matrix 【easy】
Write an efficient algorithm that searches for a value in an mx 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.
Example
Consider the following matrix:
[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return true.
Challenge
O(log(n) + log(m)) time
解法一：
1 class Solution { 2 public: 3 /* 4 * @param matrix: matrix, a list of lists of integers 5 * @param target: An integer 6 * @return: a boolean, indicate whether matrix contains target 7 */ 8 bool searchMatrix(vector<vector<int>> &matrix, int target) { 9 int row = matrix.size(); 10 if (row == 0) { 11 return false; 12 } 13 14 int col = matrix[0].size(); 15 if (col == 0) { 16 return false; 17 } 18 19 int start = 0; 20 int end = row * col - 1; 21 22 while (start + 1 < end) { 23 int mid = start + (end - start) / 2; 24 25 if (matrix[mid / col][mid % col] == target) { 26 return true; 27 } 28 else if (matrix[mid / col][mid % col] < target) { 29 start = mid; 30 } 31 else if (matrix[mid / col][mid % col] > target) { 32 end = mid; 33 } 34 } 35 36 if (matrix[start / col][start % col] == target 37 || matrix[end / col][end % col] == target) { 38 return true; 39 } 40 else { 41 return false; 42 } 43 } 44 };
注意：
1、边界值合法性检查
2、二维数组转一维数组的计算方式
解法二：
1 public class Solution { 2 public boolean searchMatrix(int[][] matrix, int target) { 3 if (matrix == null || matrix.length == 0) { 4 return false; 5 } 6 if (matrix[0] == null || matrix[0].length == 0) { 7 return false; 8 } 9 10 int row = matrix.length; 11 int column = matrix[0].length; 12 13 // find the row index, the last number <= target 14 int start = 0, end = row - 1; 15 while (start + 1 < end) { 16 int mid = start + (end - start) / 2; 17 if (matrix[mid][0] == target) { 18 return true; 19 } else if (matrix[mid][0] < target) { 20 start = mid; 21 } else { 22 end = mid; 23 } 24 } 25 if (matrix[end][0] <= target) { 26 row = end; 27 } else if (matrix[start][0] <= target) { 28 row = start; 29 } else { 30 return false; 31 } 32 33 // find the column index, the number equal to target 34 start = 0; 35 end = column - 1; 36 while (start + 1 < end) { 37 int mid = start + (end - start) / 2; 38 if (matrix[row][mid] == target) { 39 return true; 40 } else if (matrix[row][mid] < target) { 41 start = mid; 42 } else { 43 end = mid; 44 } 45 } 46 if (matrix[row][start] == target) { 47 return true; 48 } else if (matrix[row][end] == target) { 49 return true; 50 } 51 return false; 52 } 53 }
两次二分，值得借鉴！