Search a 2D Matrix

 Writean 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.

【分析-原创】

/**
 * 创建时间：2014年9月24日 上午11:26:06 项目名称：Test
 * 
 * @author Cao Yanfeng
 * @since JDK 1.6.0_21 类说明： 查找排序后的二维矩阵是否存在一个数 应用两层的二分查找，第一层是查找行
 */
public class FindMatrixTest {

	/**
	 * @param args
	 */
	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 } };
		System.out.println(searchMatrix(matrix, 13));

	}

	public static boolean searchMatrix(int[][] matrix, int target) {
		int m = matrix.length;
		int n = matrix[0].length;
		int i = 0, j = m - 1;
		while (i <= j) {
			int middleRow = i + ((j - i) >> 1);
			if (target < matrix[middleRow][0]) {
				j = middleRow - 1;
				continue;
			} else if (target > matrix[middleRow][n - 1]) {
				i = middleRow + 1;
				continue;
			} else if (target >= matrix[middleRow][0]
					&& target <= matrix[middleRow][n - 1]) {
				return find(matrix[middleRow], 0, n - 1, target);
			}
		}
		return false;
	}

	public static boolean find(int[] matrixRow, int start, int end, int target) {
		while (start <= end) {
			int middle = start + ((end - start) >> 1);
			if (matrixRow[middle] == target) {
				return true;
			} else if (matrixRow[middle] > target) {
				end = middle - 1;
			} else if (matrixRow[middle] < target) {
				start = middle + 1;
			}

		}
		return false;
	}
}

【LeetCode题目记录-13】二分搜索排序后的二维数组