Search a 2D Matrix

最新推荐文章于 2025-01-17 23:23:30 发布

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最新推荐文章于 2025-01-17 23:23:30 发布

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原文链接：http://www.cnblogs.com/gqtcgq/p/7247116.html

本文介绍了一种在有序矩阵中搜索特定值的高效算法。该矩阵的特点是每行从左到右递增排序，且每行的第一个元素大于前一行的最后一个元素。通过将二维矩阵转换为一维数组进行二分查找，实现快速定位目标值。

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.

思路：这是一个非常典型的二分查找法的应用场景，只不过是把一维数组扩展成了二维数组，因此，解题的关键是将二维数组“转换”为一维数组。代码如下：

int searchMatrix(int** matrix, int matrixRowSize, int matrixColSize, int target) 
{
	int begin = 0, end =  matrixColSize *   matrixRowSize - 1;
	int mid;
	int row, col;

	while(begin <= end)
	{
		mid = begin+(end-begin)/2;
		row = mid/matrixColSize;
		col = mid%matrixColSize;

		if(matrix[row][col] == target)	return 1;

		if(matrix[row][col] < target)	begin = mid+1;
		else	end = mid-1;
	}
	return 0;
}

转载于:https://www.cnblogs.com/gqtcgq/p/7247116.html