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Merge pull request #231 from coder-xiaomo/patch-2

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Update Time_Complexity.md
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halfrost
2022-03-01 19:02:25 -08:00
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website/content/ChapterOne/Time_Complexity.md
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@ -34,8 +34,8 @@ weight: 3
```c ```c
void hello (int n){ void hello (int n){
for( int sz = 1 ; sz < n ; sz += sz) for( int sz = 1 ; sz < n ; sz += sz )
for( int i = 1 ; i < n ; i ++) for( int i = 1 ; i < n ; i ++ )
cout << "Hello" << endl; cout << "Hello" << endl;
} }
``` ```
@ -45,7 +45,7 @@ void hello (int n){
```c ```c
bool isPrime (int n){ bool isPrime (int n){
for( int x = 2 ; x * x <= n ; x ++ ) for( int x = 2 ; x * x <= n ; x ++ )
if( n % x == 0) if( n % x == 0 )
return false; return false;
return true; return true;
} }
@ -67,7 +67,7 @@ bool isPrime (int n){
int sum( int n ){ int sum( int n ){
assert( n >= 0 ) assert( n >= 0 )
int ret = 0; int ret = 0;
for ( int i = 0 ; i <= n ; i++) for ( int i = 0 ; i <= n ; i ++ )
ret += i; ret += i;
return ret; return ret;
} }
@ -80,7 +80,7 @@ int sum( int n ){
assert( n >= 0 ) assert( n >= 0 )
if ( n == 0 ) if ( n == 0 )
return 0; return 0;
return n + sum( n - 1); return n + sum( n - 1 );
} }
``` ```
@ -96,14 +96,14 @@ int sum( int n ){
```c ```c
int binarySearch(int arr[], int l, int r, int target){ int binarySearch(int arr[], int l, int r, int target){
if( l > r) if( l > r )
return -1; return -1;
int mid = l + (r-l)/2;//防溢出 int mid = l + ( r - l ) / 2; // 防溢出
if(arr[mid] == target) if(arr[mid] == 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);
} }
@ -119,16 +119,16 @@ int binarySearch(int arr[], int l, int r, int target){
```c ```c
int f(int n){ int f(int n){
assert( n >= 0 ); assert( n >= 0 );
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。
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