Search a 2D Matrix | & II

Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.

This matrix has the following properties:

• Integers in each row are sorted from left to right.
• Integers in each column are sorted from up to bottom.
• No duplicate integers in each row or column.

Example

Consider the following matrix:

[
  [1, 3, 5, 7],
  [2, 4, 7, 8],
  [3, 5, 9, 10]
]

Given target = 3, return 2.

分析：

因为数组里的数有上面三个特性，所以我们可以从左下角开始找。如果当前值比target大，明显往上走，比target小，往右走 (其它角没有这样的性质)，如果一样斜上走。Time complexity: (O(m + n)) (m = matrix.length, n = matrix[0].length)

public class Solution {
    public int searchMatrix(int[][] matrix, int target) {
        // check corner case
        if (matrix == null || matrix.length == 0 || matrix[].length == 0) {
            return 0;
        }     
        // from bottom left to top right
        int x = matrix.length - 1;
        int y = 0;
        int count = 0;
        
        while (x >= 0 && y < matrix[0].length) {
            if (matrix[x][y] < target) {
                y++;
            } else if (matrix[x][y] > target) {
                x--;
            } else {
                count++;
                x--;
                y++;
            }
        }
        return count;
        
    }
}

Search a 2D Matrix I

Write an efficient algorithm that searches for a value in an mx 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.

Example

Consider the following matrix:

[
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]

Given target = 3, return true.

分析：

根据数组的特点，我们需要找出一个大范围（row），然后再确定小范围。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
        int rowCount = matrix.length, colCount = matrix[0].length;
        if (matrix[0][0] > target || matrix[rowCount - 1][colCount - 1] < target) return false;
        
        int row = getRowNumber(matrix, target);
        int column = getColumnNumber(matrix, row, target);
        
        return matrix[row][column] == target;
        
    }

    public int getRowNumber(int[][] matrix, int target) {
        int start = 0, end = matrix.length - 1, width = matrix[0].length;
        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid][width - 1] == target) {
                return mid;
            } else if (matrix[mid][width - 1] > target) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        } 
        return start;
    }
    
    public int getColumnNumber(int[][] matrix, int row, int target) {
        int start = 0, end = matrix[0].length;
        
        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (matrix[row][mid] == target) {
                return mid;
            } else if (matrix[row][mid] > target) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        } 
        return start;
    }
}