Search a 2D Matrix

Search a 2D Matrix

问题：

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.

思路：

　　二分查找

我的代码：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0)   return false;
        int row = matrix.length;
        int col = matrix[0].length;
        int[] cols = new int[row];
        for(int i = 0; i < row; i++)
        {
            cols[i] = matrix[i][0];
        }
        int rowIndex = getIndex(cols, target);
        if(rowIndex < 0)    return false;
        int[] rows = new int[col];
        for(int i = 0; i < col; i++)
        {
            rows[i] = matrix[rowIndex][i];
        }
        int colIndex = getIndex(rows, target);
        return rows[colIndex] == target ? true : false;
    }
    public int getIndex(int[] array, int target)
    {
        int left = 0;
        int right = array.length - 1;
        while(left <= right)
        {
            int mid = (left + right)/2;
            if(target == array[mid])
            {
               return mid; 
            }
            else if(target > array[mid])
            {
                left = mid + 1;
            }
            else
            {
                right = mid - 1;
            }
        }
        return left - 1;
    }
}

他人代码：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        
        int rows = matrix.length;
        int cols = matrix[0].length;
        
        int num = rows * cols;
        
        int left = 0;
        int right = num - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            int row = mid / cols;
            int col = mid % cols;
            
            int n = matrix[row][col];
            
            if (n == target) {
                return true;
            } else if (n < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return false;        
    }
}

学习之处：

　　他人代码的思路更加简洁，把整个矩阵看成一个Array，好想法，rowNum = num/row colNum = num % col 思路实在是妙