題目：

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, return true.

Given target = 20, return false.

題解：

Solution 1 ()

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        if(m == 0) return false;
        int n = matrix[0].size();

        for(int
 
 c=0; c<n; ++c) {
            if(matrix[m-1][c] < target) continue;
            if(matrix[m-1][c] == target || matrix[0][c] == target) return true;
            for(int r=m-2; r>=0; --r) {
                if(matrix[r][c] == target) return true;
                if(matrix[r][c] > target) continue
 
;
                if(matrix[r][n-1] < target) return false;
                for(int k=c; k<n; ++k) {
                    if(matrix[r][k] == target) return true;
                    if(matrix[r][k] < target) continue;
                    break;
                }    
            }  
        }
        return false; 
    }
};

　　　從右上角開始，如果數字小於target，所在行不可能有target這個數，如果數字大於target，所在列不可能有target這個數（也可以從左下角開始）

Solution 2 ()

class Solution {
public:
    bool searchMatrix(vector<vector<int>> &matrix, int target) {
        if (matrix.empty() || matrix[0].empty()) return false;
        int m = matrix.size();
        int n = matrix[0].size();
        for (int i = 0, j = n - 1; i < m && j >= 0;) {
            if (matrix[i][j] == target) return true;
            if (matrix[i][j] > target) j--;
            else i++;
        }
        return false;
    }
};

【LeetCode】240. Search a 2D Matrix II