74 Search a 2D Matrix

74. Search a 2D Matrix

1. Question

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

2. Implementation

(1) Binary Search

思路: 因为每一行的第一个数肯定比上一行的最后一个数大，所以matrix[0][0]到matrix[m - 1][n - 1]是严格递增的, 可以直接把matrix降成一维数组，对这个一维数组进行二分查找

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

        int m = matrix.length, n = matrix[0].length;
        int start = 0, end = m * n - 1, mid = 0;

        while (start <= end) {
            mid = start + (end - start) / 2;
            int row = mid / n;
            int col = mid % n;

            if (matrix[row][col] == target) {
                return true;
            }
            else if (matrix[row][col] < target) {
                start = mid + 1;
            }
            else {
                end = mid - 1;
            }
        }
        return false;
    }
}

3. Time & Space Complexity

时间复杂度O(logmn), m为matrix的行数, row为matrix的列数，空间复杂度O(1)