leetcode oj java Search a 2D Matrix

一、问题描述：

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, return true.

二、解决方案：

矩阵行扫描是有序的，用典型的二分查找。矩阵为m*n, 看为一个1 * （m*N）的数组，第k个位置对应到矩阵的[k/n][k%n] 

三、代码：

package T12;/** * @author 作者 : xcy * @version 创建时间：2016年12月22日 下午11:48:33 *          类说明 */public class t74 {    public static void main(String[] args) {        // TODO Auto-generated method stub        int[][] matrix = { { 1, 3, 5, 7 }, { 10, 11, 16, 20 }, { 23, 30, 34, 50 } };        int target = 51;        System.out.println(searchMatrix(matrix, target));    }    public static boolean searchMatrix(int[][] matrix, int target) {        boolean flag = false;        int m = matrix.length;        if(m==0){            return false;        }        int n = matrix[0].length;        int num = m * n;        int start = 0;        int end = num;        int x = 0;        int y = 0;        int mid = 0;        while (start < end) {            mid = (start + end) / 2;            x = mid / n;            y = mid % n;            if (target == matrix[x][y]) {                return true;            }            if (target < matrix[x][y]) {                end = start + (end - start) / 2;            }            if (target > matrix[x][y]) {                start = end - (end - start) / 2;            }        }        return flag;    }}