LeetCode 刷題記錄74. Search a 2D Matrix

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 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],yiweide
[23, 30, 34, 50]
]
target = 13
Output: false
解法1：
二分法
將一維的二分法映射到二維的數組中
假設一維的下標爲mid，則映射到二維數組的行和列下標爲matrix[mid / n][mid % n]，n爲列的長度
c++：

 if len(matrix) == 0 or len(matrix[0]) == 0: return False
        m , n = len(matrix), len(matrix[0])
        left , right = 0 , m * n - 1 
        while left <= right:
            mid = left + (right - left) / 2
            if matrix[mid / n][mid % n] == target: return True
            elif matrix[mid / n][mid % n] < target:
                left = mid + 1
            else:
                right = mid - 1
            
        
        return False

java：

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length == 0 || matrix[0].length == 0) return false;
        int m = matrix.length;
        int n = matrix[0].length;
        int left = 0;
        int right = m * n - 1;
        while(left <= right){
            int mid = left + (right - left) / 2;
            if(matrix[mid / n][mid % n] == target) return true;
            else if(matrix[mid / n][mid % n] < target){
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return false;
    }
}

python：

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if len(matrix) == 0 or len(matrix[0]) == 0: return False
        m , n = len(matrix), len(matrix[0])
        left , right = 0 , m * n - 1 
        while left <= right:
            mid = left + (right - left) / 2
            if matrix[mid / n][mid % n] == target: return True
            elif matrix[mid / n][mid % n] < target:
                left = mid + 1
            else:
                right = mid - 1
            
        
        return False

解法2：
二分法變體
將二分法的區間改爲前閉後開，則初始值則爲m * n而不是m\ * n-1,循環的條件爲小於， right = mid;而不是 right = mid-1;

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.size() == 0 || matrix[0].size() == 0) return false;
        int m = matrix.size();
        int n = matrix[0].size();
        int left = 0;
        int right = m * n;
        while(left < right){
            int mid = left + (right - left) / 2;
            if(matrix[mid / n][mid % n] == target) return true;
            else if(matrix[mid / n][mid % n] < target){
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return false;
    }
};