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;

    }

}

View Code

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;        

    }

}

View Code

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