search-a-2d-matrix

packagecom.ytx.array;

/** 题目：search-a-2d-matrix

 *  描述：

    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.

             For example,

             Consider the following matrix:

             [

                    [1,   3,  5,  7],

                    [10, 11, 16, 20],

                    [23, 30, 34, 50]

             ]

             Given target =3,returntrue.

 *@authoryuantian xin

 * 

 * 写一个高效的算法在一个m乘n的矩阵中查找一个值，找到了返回true。这个m行n列矩阵的特点是，每一行数的大小是递增的，

 * 每一列数的大小也是递增的。

 *

 * 换种思路，以右上角或者左下角为基点：

 * 例如以右上角为基点，往左递减，往下递增。如果要查找的数字target比右上角的数字matrix[0][col-1]大就往下找,i++,

 * 如果比matrix[0][col-1]小就往左找，j--,循环直到找到为止。

 *

 * 另外一种解法：二分查找，把每一行当做一个有序的一维数组进行二分查找。

 *

 *  public boolean searchMatrix(int[][] matrix,int target) {

       int row = matrix.length;

       int col = matrix[0].length;

        

        for(int i=0;i < row;i++){

            if(target < = matrix[i][col-1]){

               int start = 0, end = col - 1;

                while( start <= end ) {

                   int mid=(start + end) / 2;

                    if(target == matrix[i][mid])

                        return true;

                    else if( target< matrix[i][mid])

                       end = mid- 1;

                     else

                       start = mid + 1;

                }

            }

        }

        return false;

    }

 *

 *

 */

publicclass Search_a_2d_matrix {

   

       publicbooleansearchMatrix(int[][]matrix,int target) {

             

             introw = matrix.length;

             intcol = matrix[0].length;

             booleanflag = false;

             //从右上角开始找，往左边数字递减，往下数字递增

             inti = 0, j = col - 1;

             while(i < row && j >= 0 ) {

                    

                    if(target== matrix[i][j]) {

                           flag= true;

                           break;

                    }else if(target< matrix[i][j]) {

                           j--;

                    }else {

                           i++;

                    }

             }

       return flag;

    }

       

       publicstatic void main(String[]args) {

             int[][]matrix = {{1,3,5,7},{10,11,16,20},{23,30,34,50}};

             booleanbn = newSearch_a_2d_matrix().searchMatrix(matrix, 90);

             System.out.println(bn);

       }

}