Problem:
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
思路：
将二维矩阵转化为一位数组，使用二分查找即可。
转换公式：
(m imes n)维矩阵转换为数组：matrix[x][y] (Rightarrow) a[x*n+y];
数组转换为(m imes n)维矩阵：a[x] (Rightarrow) matrix[x/n][x%n]；
Solution:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.size() == 0 || matrix[0].size() == 0)
return false;
int m = matrix.size(), n = matrix[0].size();
/*注意一定要先判断矩阵的行和列是否为0，否则会报错：runtime error: reference binding to null pointer of type 'struct value_type' (stl_vector.h)
产生此问题的原因是若列为0，则matrix[0]无法引用*/
int l = 0, r = m * n - 1;
while (l != r) {
int mid = l + (r - l) / 2;
if (matrix[mid/n][mid%n] < target)
l = mid + 1;
else
r = mid;
}
return matrix[r/n][r%n] == target;
}
性能：
Runtime: 8 ms Memory Usage: 9.9 MB
