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题目地址：

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.

既然有序，那毫无疑问，二分查找肯定是跑不了了，首先垂直方向二分查找，确定好target所在的行，然后水平方向二分查找。

public class Search2DMatrix {
       public static boolean searchMatrix(int[][] matrix, int target) {        if (matrix == null)            return false;        if (matrix.length == 0)            return false;        int elemCnt = 0;        for (int i = 0; i < matrix.length; i++) {            elemCnt += matrix[i].length;        }        if (elemCnt == 0)            return false;        if (target < matrix[0][0])            return false;        if (target > matrix[matrix.length - 1][matrix[0].length - 1])            return false;        int top = 0;        int bottom = matrix.length - 1;        int midRow = top + (bottom - top) / 2;        while (top <= bottom) {            if (matrix[midRow][0] == target) {                return true;            } else if (matrix[midRow][0] > target) {                bottom = midRow - 1;                midRow = top + (bottom - top) / 2;                continue;            } else {                top = midRow + 1;                midRow = top + (bottom - top) / 2;                continue;            }        }        if (top < 0)            top = 0;        if (top > matrix.length - 1)            top = matrix.length - 1;        if (matrix[top][0] > target)            top--;        int left = 0;        int right = matrix[top].length - 1;        int midCol = left + (right - left) / 2;        while (left <= right) {            if (matrix[top][midCol] == target) {                return true;            } else if (matrix[top][midCol] > target) {                right = midCol - 1;                midCol = left + (right - left) / 2;            } else {                left = midCol + 1;                midCol = left + (right - left) / 2;            }        }        return false;    }    public static void main(String[] args) {        int[][] maxtrix = new int[][]{                {
  0, 1, 2, 3, 5},                {
  6, 8, 9, 10, 11},                {
  12, 13, 14, 16, 17},                {
  18, 19, 20, 21, 23},                {
  24, 26, 27, 28, 29},                {
  30, 31, 32, 33, 34},                {
  36, 37, 38, 39, 40},                {
  42, 44, 45, 46, 47}        };        System.out.println(searchMatrix(maxtrix, 41));        return;    }}

最后注意一下这种特殊情况的矩阵：{

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