先判断数组是否为空，为空返回false
不为空进入循环
获得矩阵的行数：rows=matrix.length
获得矩阵的列数：columns=matrix[0].length
从最右上角开始找，初始化row和column:row=0,column=columns-1;
所以while条件为 row<rows&&column>=0
如果找到，即matrix[row][column]=number，并将found置为true
如果没找到，当前数大于Number，在左边找，column–
当前数小于Number，在右边找，row++
package findInMatrix;
public class FindInMatrix {
public static void main(String[] args) {
// TODO Auto-generated method stub
test1();
test2();
}
static void test1() {
int matrix[][] = { { 1, 2, 8, 9 }, { 2, 4, 9, 12 }, { 4, 7, 10, 13 },
{ 6, 8, 11, 15 } };
System.out.println(find(matrix, 1));
}
static void test2() {
System.out.println(find(null, 7));
}
static boolean find(int[][] matrix, int number) {
boolean found = false;
if (matrix != null) {
int rows = matrix.length;
int columns = matrix[0].length;
int row = 0;
int column = columns – 1;
while (row < rows && column >= 0) {
if (matrix[row][column] == number) {
found = true;
break;
} else if (matrix[row][column] > number) {
–column;
} else {
++row;
}
}
}
return found;
}
}
