Search a 2D Matrix

题目原型：

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.

基本思路：

其实就是二分查找的变形。

	public boolean searchMatrix(int[][] matrix, int target)
	{
		if(matrix.length>0)
			return byBiSearch(matrix, target, 0, matrix.length*matrix[0].length-1);
		else
			return false;
	}
	
	
	//二分查找
	public boolean byBiSearch(int[][] matrix, int target,int begin , int end)
	{
		if(begin>end)
			return false;
		int mid = (begin+end)/2;
		int row = mid/matrix[0].length;
		int column = mid%matrix[0].length;
		if(matrix[row][column]==target)
			return true;
		else if(matrix[row][column]>target)
			return byBiSearch(matrix, target, begin, mid-1);
		else
			return byBiSearch(matrix, target, mid+1, end);
	}

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Search a 2D Matrix

原文：http://blog.csdn.net/cow__sky/article/details/21038403