74. Search a 2D Matrix

Write an efficient algorithm that searches for a value in anm 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, returntrue.

全部有序，使用二分查詢，關鍵是二維座標和一維座標的轉換

// 時間複雜度是O(logm+logn)，空間上只需兩個輔助變數，因而是O(1)
	public boolean searchMatrix(int[][] matrix, int target) {
		if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
			return false;

		int m = matrix.length;
		int n = matrix[0].length;
		int start = 0;
		int end = m * n - 1;
		while (start <= end) {
			int mid = start + (end - start) / 2;
			int midX = mid / n;
			int midY = mid % n;

			if (matrix[midX][midY] == target)
				return true;

			if (matrix[midX][midY] < target) {
				start = mid + 1;
			} else {
				end = mid - 1;
			}
		}

		return false;
	}
 

240. Search a 2D Matrix II

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.

此種方法同樣適用於上一題目

// 時間複雜度是O(m+n)
	// 從右上角開始, 比較target 和 matrix[row][col]的值.
	// 如果小於target, 則該行不可能有此數, 所以row++;
	// 如果大於target, 則該列不可能有此數, 所以col--.
	// 遇到邊界則表明該矩陣不含target.
	public boolean searchMatrix1(int[][] matrix, int target) {
		if (matrix == null || matrix.length == 0)
			return false;
		int rows = matrix.length;
		int cols = matrix[0].length;
		int row = 0;
		int col = cols - 1;
		while (row < rows && col >= 0) {
			if (matrix[row][col] == target)
				return true;
			else if (matrix[row][col] < target)
				row++;

			else
				col--;
		}
		return false;
	}