74. Search a 2D Matrix

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.

矩阵每行从左往右是有序的，每列从上往下是有序的，那就可以用二分搜索。程序如下所示：

class Solution {    public boolean searchMatrix(int[][] matrix, int target) {        if (matrix.length == 0||matrix[0].length == 0){            return false;        }        int row = matrix.length, col = matrix[0].length;         int left = 0, right = row - 1, mid = 0;        while (left <= right){            mid = left + (right - left)/2;            if (matrix[mid][0] < target){                if (mid+1 <= right&&matrix[mid+1][0] > target){                    break;                }                left = mid + 1;            }            else if (matrix[mid][0] > target){                right = mid - 1;            }            else {                return true;            }        }        int index = mid;        left = 0;        right = col - 1;        while (left <= right){            mid = left + (right - left)/2;            if (matrix[index][mid] < target){                left = mid + 1;            }            else if (matrix[index][mid] > target){                right = mid - 1;            }            else {                return true;            }        }        return false;    }}