题目英文

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 from left to right.
The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

题目中文

编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：

每行中的整数从左到右按升序排列。
每行的第一个整数大于前一行的最后一个整数。

示例 1:

输入:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
输出: true

示例 2:

输入:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
输出: false

算法实现

public class Solution
{
    public bool SearchMatrix(int[][] matrix, int target)
    {
        if (matrix == null)
            return false;

        int rows = matrix.Length;
        for (int i = 0; i < rows; i++)
        {
            if (matrix[i] == null || matrix[i].Length == 0)
                return false;

            int j = matrix[i].Length - 1;
            if (matrix[i][j] == target)
                return true;
            if (matrix[i][j] > target)
                return BinSearch(matrix[i], target);
        }
        return false;
    }

    public bool BinSearch(int[] key, int target)
    {
        int left = 0;
        int right = key.Length;
        while (left <= right)
        {
            int mid = (left + right)/2;
            if (key[mid] == target)
                return true;
            if (key[mid] < target)
                left = mid + 1;
            else
                right = mid - 1;
        }
        return false;
    }
}