My code:

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
            return false;
        for (int i = 0; i < matrix.length; i++) {
            int begin = 0;
            int end = matrix[0].length - 1;
            while (begin <= end) {
                int middle = begin + (end - begin) / 2;
                if (matrix[i][middle] > target) {
                    end = middle - 1;
                }
                else if (matrix[i][middle] < target) {
                    begin = middle + 1;
                }
                else
                    return true;
            }
        }
        return false;
    }
}

我觉得这道题木就是每一行用一次binary seach，O(n log m), 没什么难的。

Anyway, Good luck, Richardo!

My code:

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        
        int i = 0;
        int j = matrix[0].length - 1;
        while (i < matrix.length && j >= 0) {
            int curr = matrix[i][j];
            if (curr == target) {
                return true;
            }
            else if (curr > target) {
                j--;
            }
            else {
                i++;
            }
        }
        return false;
    }
}

reference:
https://discuss.leetcode.com/topic/20064/my-concise-o-m-n-java-solution/2

time complexity: O(m + n)

这个做法，并没有很好地利用 binary search

然后 binary search 通俗的做法是下面这个链接：
https://discuss.leetcode.com/topic/19540/ac-java-solution-with-divide-and-conquer-follow-the-hide-tags/2

复杂度是 O(row * log(col))
具体思路就是，先找到 <= target 的所有行
然后分别进行 binary search

然后看了下我以前的做法，直接对所有行进行 Binary search,
其实在时间复杂度上没有区别。

Anyway, Good luck, Richardo! — 09/21/2016