240. Search a 2D Matrix II

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本文介绍了一种高效的算法，用于在一个m x n的二维矩阵中查找特定值。该矩阵具有每一行从左到右递增排序，每一列从上到下递增排序的特点。文章通过一个具体的例子详细阐述了如何实现这一算法，并给出了一个O(M+N)复杂度的解决方案。

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

First read discussions

http://articles.leetcode.com/searching-2d-sorted-matrix

http://articles.leetcode.com/searching-2d-sorted-matrix-part-ii

https://leetcode.com/discuss/51108/is-theres-a-o-log-m-log-n-solution-i-know-o-n-m-and-o-m-log-n

Solution 1

// O(M + N) Divide and conquer
	public static boolean searchMatrix(int[][] matrix, int target) {
		if (matrix == null || matrix.length < 1 || matrix[0].length < 1) {
			return false;
		}
		int col = matrix[0].length - 1;
		int row = 0;
		while (col >= 0 && row <= matrix.length - 1) {
			if (target == matrix[row][col]) {
				return true;
			} else if (target < matrix[row][col]) {
				col--;
			} else if (target > matrix[row][col]) {
				row++;
			}
		}
		return false;
	}