240 Search a 2D Matrix II

240. Search a 2D Matrix II

1. Question

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, returntrue.

Given target=20, returnfalse.

2. Implementation

(1) Binary Search

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }

        int m = matrix.length, n = matrix[0].length;

        int rowStart = 0, rowEnd = m - 1, rowMid = 0;

        while (rowStart + 1 < rowEnd) {
            rowMid = rowStart + (rowEnd - rowStart) / 2;

            if (matrix[rowMid][0] < target) {
                rowStart = rowMid + 1;
            }
            else {
                rowEnd = rowMid;
            }
        }

        rowStart = matrix[rowStart][0] >= target ? rowStart : rowEnd;

        for (int i = 0; i <= rowStart; i++) {
            int colStart = 0, colEnd = n - 1, colMid = 0;

            while (colStart <= colEnd) {
                colMid = colStart + (colEnd - colStart) / 2;

                if (matrix[i][colMid] == target) {
                    return true;
                }
                else if (matrix[i][colMid] < target) {
                    colStart = colMid + 1;
                }
                else {
                    colEnd = colMid - 1;
                }
            }
        }
        return false;
    }
}

3. Time & Space Complexity

Binary Search: 时间复杂度O(logm + mlogn), 空间复杂度O(1)