mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-05 16:36:41 +08:00
Merge pull request #231 from coder-xiaomo/patch-2
Update Time_Complexity.md
This commit is contained in:
@ -103,7 +103,7 @@ int binarySearch(int arr[], int l, int r, int target){
|
|||||||
return mid;
|
return mid;
|
||||||
else if (arr[mid] > target)
|
else if (arr[mid] > target)
|
||||||
return binarySearch(arr,l,mid-1,target);
|
return binarySearch(arr,l,mid-1,target);
|
||||||
eles
|
else
|
||||||
return binarySearch(arr,mid+1,r,target);
|
return binarySearch(arr,mid+1,r,target);
|
||||||
}
|
}
|
||||||
|
|
||||||
@ -122,13 +122,13 @@ int f(int n){
|
|||||||
if( n == 0 )
|
if( n == 0 )
|
||||||
return 1;
|
return 1;
|
||||||
return f( n - 1 ) + f ( n - 1 );
|
return f( n - 1 ) + f ( n - 1 );
|
||||||
|
}
|
||||||
```
|
```
|
||||||
|
|
||||||
上述这次递归调用的次数为 2^0^ + 2^1^ + 2^2^ + …… + 2^n^ = 2^n+1^ - 1 = O(2^n)
|
上述这次递归调用的次数为 2^0^ + 2^1^ + 2^2^ + …… + 2^n^ = 2^n+1^ - 1 = O(2^n)
|
||||||
|
|
||||||
|
|
||||||
> 关于更加复杂的递归的复杂度分析,请参考,主定理。主定理中针对各种复杂情况都给出了正确的结论。
|
> 关于更加复杂的递归的复杂度分析,请参考主定理。主定理中针对各种复杂情况都给出了正确的结论。
|
||||||
|
|
||||||
|
|
||||||
----------------------------------------------
|
----------------------------------------------
|
||||||
|
Reference in New Issue
Block a user