Search a 2D Matrix,在有序矩阵查找，二分查找的变形; 行有序，列有序查找。

问题描述：矩阵每一行有序，每一行的最后一个元素小于下一行的第一个元素，查找。

算法分析：这样的矩阵其实就是一个有序序列，可以使用折半查找算法。

public class SearchInSortedMatrix 
{
	public static boolean searchMatrix(int[][] matrix, int target)
	{
		int m = matrix.length;
		int n = matrix[0].length;
		int low = 0;
		int high = m*n - 1;
		while(low <= high)
		{
			int mid = (low + high)/2;
			int row = mid / n;
			int column = mid % n;
			if(matrix[row][column] == target)
			{
				return true;
			}
			else if(matrix[row][column] < target)
			{
				low = mid + 1;
			}
			else
			{
				high = mid - 1;
			}
		}
		return false;
	}
}

 

问题描述：二维矩阵行有序，列有序，进行查找。

算法分析：有两种方法，一种是将矩阵按中心点分成左上，左下，右上，右下，四部分，进行递归查找。

　　　　　还有一种比较巧妙的查找方法，就是从左下角或者右上角的元素进行查找。

	//矩阵每一行有序，每一列有序
	public  boolean searchMatrix2(int[][] matrix, int target)
	{
		int m = matrix.length;
		int n = matrix[0].length;
		return helper(matrix, 0, m-1, 0, n-1, target);
	}
	public boolean helper(int[][] matrix, int rowStart, int rowEnd, int colStart, int colEnd, int target)
	{
		int rm = (rowStart + rowEnd)/2;
		int cm = (colStart + colEnd)/2;
		if(rowStart > rowEnd || colStart > colEnd)
		{
			return false;
		}
		if(matrix[rm][cm] == target)
		{
			return true;
		}
		else if(matrix[rm][cm] > target)
		{
			return helper(matrix, rowStart, rm - 1, colStart, cm - 1, target)
				|| helper(matrix, rm, rowEnd, colStart, cm - 1, target)
				|| helper(matrix, rowStart, rm - 1, cm, colEnd, target);
		}
		else
		{
			return helper(matrix, rm + 1, rowEnd, cm + 1, colEnd, target)
				|| helper(matrix, rm + 1, rowEnd, colStart, cm, target)
				|| helper(matrix, rowStart, rm, cm + 1, colEnd, target);
				   
		}
	}
	
	//从右上角元素进行查找
	public boolean searchMatrix3(int[][] matrix, int target)
	{
		int m = matrix.length;
		int n = matrix[0].length;
		int low = 0;
		int high = m*n - 1;
		while(low <= high)
		{
			int mid = (low + high)/2;
			int row = mid / n;
			int column = mid % n;
			if(matrix[row][column] == target)
			{
				return true;
			}
			else if(matrix[row][column] < target)
			{
				low = mid + 1;
			}
			else
			{
				high = mid - 1;
			}
		}
		return false;
	}