【Lintcode】028.Search a 2D Matrix

题目：

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.

题解：

Solution 1 ()

class Solution {
public:
    /**
     * @param 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) {
        if (matrix.empty() || matrix[0].empty()) {
            return false;
        }
        
        int rowEnd = matrix.size() - 1;
        int colEnd = matrix[0].size() - 1;
        int start = 0, mid = 0;
        //rowMid
        while (start + 1 < rowEnd) {
            mid = start + (rowEnd - start) / 2;
            if (matrix[mid][0] == target) {
                return true;
            } else if (matrix[mid][0] > target) {
                rowEnd = mid;
            } else {
                start = mid;
            }    
        }
        
        int row = 0; 
        if (matrix[start][0] <= target && matrix[start][colEnd] >= target) {
            row = start;
        } else if (matrix[rowEnd][0] <= target && matrix[rowEnd][colEnd] >= target) {
            row = rowEnd;
        } else {
            return false;
        }
        start = 0;
        
        //colMid
        while (start + 1 < colEnd) {
            mid = start + (colEnd - start) / 2;
            if (matrix[row][mid] == target) {
                return true;
            } else if (matrix[row][mid] > target) {
                colEnd = mid;
            } else {
                start = mid;
            }    
        }
        
        if(matrix[row][start] == target || matrix[row][colEnd] == target) {
            return true;
        }
                
        return false;
    }
};

Solution 2 ()

class Solution {
public:
    /**
     * @param 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) {
        if (matrix.empty() || matrix[0].empty()) {
            return false;
        }
        
        int m = matrix.size(), n =  matrix[0].size();
        int start = 0;
        int end = m * n - 1;
        int mid = 0;
        
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            int row = mid / n;
            int col = mid % n;
            if (matrix[row][col] == target) {
                return true;
            } else if (matrix[row][col] > target) {
                end = mid;
            } else {
                start = mid;
            } 
        }
        
        if (matrix[start / n][start % n] == target) {
            return true;
        } else if (matrix[end / n][end % n] == target) {
            return true;
        } else {
            return false;
        }
    }
};