leetcode 74. Search a 2D Matrix (二分)
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
C++
class Solution {
public:
bool searchMatrix(vector<vector<int>>& a, int t) {
int n=a.size();
if(n==0)return false;
int m=a[0].size();
if(m==0)return false;
int l=0,r=n*m-1;
while(l<r){
int k=l+(r-l)/2;
if(a[k/m][k%m]>=t)r=k;
else l=k+1;
}
return a[r/m][r%m]==t;
}
};
Java
class Solution {
public boolean searchMatrix(int[][] a, int t) {
int n=a.length;
if(n==0)return false;
int m=a[0].length;
if(m==0)return false;
int l=0,r=n*m-1;
while(l<r){
int k=l+(r-l)/2;
if(a[k/m][k%m]>=t)r=k;
else l=k+1;
}
return a[r/m][r%m]==t;
}
}
Python
class Solution:
def searchMatrix(self, a: List[List[int]], t: int) -> bool:
n=len(a)
if n==0:
return False
m=len(a[0])
if m==0:
return False
l,r=0,n*m-1
while l<r:
k=l+(r-l)//2
if a[k//m][k%m]>=t:
r=k
else:
l=k+1
return a[r//m][r%m]==t
Go
func searchMatrix(a [][]int, t int) bool {
n:=len(a)
if n==0 {
return false
}
m:=len(a[0])
if m==0 {
return false
}
l,r:=0,n*m-1
for l<r {
k:=l+(r-l)/2
if a[k/m][k%m]>=t {
r=k
} else {
l=k+1
}
}
return a[r/m][r%m]==t
}