28. Search a 2D Matrix 【easy】

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：

class Solution {
public:
    /*
     * @param matrix: matrix, a list of lists of integers
     * @param target: An integer
     * @return: a boolean, indicate whether matrix contains target
     */
    bool searchMatrix(vector<vector<int>> &matrix, int target) {
        int row = matrix.size();
        if (row == 0) {
            return false;
        }
        
        int col = matrix[0].size();
        if (col == 0) {
            return false;
        }
        
        int start = 0;
        int end = row * col - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            
            if (matrix[mid / col][mid % col] == target) {
                return true;
            }
            else if (matrix[mid / col][mid % col] < target) {
                start = mid;
            }
            else if (matrix[mid / col][mid % col] > target) {
                end = mid;
            }
        }
        
        if (matrix[start / col][start % col] == target 
            || matrix[end / col][end % col] == target) {
            return true;
        }
        else {
            return false;
        }
    }
};

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：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) {
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0) {
            return false;
        }
        
        int row = matrix.length;
        int column = matrix[0].length;
        
        // find the row index, the last number <= target 
        int start = 0, end = row - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid][0] == target) {
                return true;
            } else if (matrix[mid][0] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (matrix[end][0] <= target) {
            row = end;
        } else if (matrix[start][0] <= target) {
            row = start;
        } else {
            return false;
        }
        
        // find the column index, the number equal to target
        start = 0;
        end = column - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[row][mid] == target) {
                return true;
            } else if (matrix[row][mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (matrix[row][start] == target) {
            return true;
        } else if (matrix[row][end] == target) {
            return true;
        }
        return false;
    }
}

两次二分，值得借鉴！