面試題11:數值的整除次方
題目:實現函數double Power(double base,int exponent),求base的exponent次方。不得使用庫函數,同時不需要考慮大數問題。
解析:當指數爲負數的時候,可以先對指數求絕對值,然後算出次方的結果之後再取倒數。既然有求倒數,我們很自然就要想到有沒有可能對0求倒數。當底數(base)爲0時指數是負數的時候,如果不做特殊處理,就會出現對0求倒數從而導致程序運行錯誤。
bool g_InvalidInput = false;
double Power(double base,int exponent)
{
g_InvalidInput = false;
if(equal(base,0.0) && exponent < 0)
{
g_InvalidInput = true;
return 0.0;
}
unsigned int absExponent = (unsigned int)(exponent);
if(exponent < 0)
absExponent = (unsigned int)(-exponent);
double result = PowerWithUnsignedExponent(base,absExponent);
if(exponent < 0)
result = 1.0/result;
return result;
}
double PowerWithUnsignedExponent(double base,unsigned int exponent)
{
double result = 1.0;
for(int i=1;i <= exponent; ++i)
result*=base;
return result;
}
bool equal(double num1,double num2)
{
if((num1-num2>-0.0000001)
&&(num1-num2<0.0000001))
return true;
else
return false;
}
全面又高效的解法,確保我們能拿到Offer
我們可以用如下公式求a的n次方:
double PowerWithUnsignedExponent(double base,unsigned int exponent)
{
if(exponent == 0)
return 1;
if(exponent == 1)
return base;
double result = PowerWithUnsignedExponent(base,exponent>>1);
result *= result;
if(exponent * 0x1 ==1)
result*=base;
return result;
}
我們用右移運算符代替了除以2,用位與運算符代替求餘運算符(%)來判斷一個數是奇數還是偶數。位運算的效率比乘除法及求餘運算的效率要高很多。既然要優化代碼,我們就把優化做到極致。
面試題12:打印1到最大的n位數
題目:輸入數字n,按順序打印出從1醉倒的n位十進制數。比如輸入3,則打印出1、2、3一直到最大的3位數即999。
void Print1ToMaxOfNDigits(int n)
{
if(n <= 0)
return;
char* number = new char[n+1];
memset(number,'0',n);
number[n] = '\0';
while(!Increment(number))
{
PrintNumber(number);
}
delete []number;
}
bool Increment(char* number)
{
bool isOverflow = false;
int nTakeOver = 0;
int nLength = strlen(number);
for(int i=nLength-1;i>=0;i--)
{
int nSum = number[i]-'0'+nTakeOver;
if(i == nLength -1)
nSum ++;
if(nSum >= 10)
{
if(i == 0)
isOverflow = true;
else
{
nSum -= 10;
nTakeOver = 1;
number[i] = '0'+nSum;
}
}
else
{
number[i]='0'+nSum;
break;
}
}
return isOverflow;
}
void PrintNumber(char* number)
{
bool isBeginning0 = true;
int nLength = strlen(number);
for(int i=0;i<nLength;++i)
{
if(isBeginning0 && number[i] != '0')
isBeginning0 = false;
if(!isBeginning0)
{
printf("%c",number[i]);
}
}
printf("\t");
}
面試題13:在O(1)時間刪除鏈表結點
題目:給定單向鏈表的頭指針和一個結點指針,定義一個函數在O(1)時間刪除該結點。鏈表結點與函數的定義如下:
struct ListNode
{
int m_nValue;
ListNode* m_pNext;
};
void DeleteNode(ListNode** pListHead,ListNode* pToBeDeleted)
{
if(!pListHead || !pToBeDeleted)
return;
//要刪除的結點不是尾結點
if(pToBeDeleted->m_pNext != NULL)
{
ListNode* pNext = pToBeDeleted->m_pNext;
pToBeDeleted->m_nValue = pNext->m_nValue;
pToBeDeleted->m_pNext = pNext->m_pNext;
delete pNext;
pNext = NULL;
}
//鏈表只有一個結點,刪除頭結點(也是尾結點)
else if(*pListHead == pToBeDeleted)
{
delete pToBeDeleted;
pToBeDeleted = NULL;
*pListHead = NULL;
}
//鏈表中有多個結點、刪除尾結點
else
{
ListNode* pNode =*pListHead;
while(pNode->m_pNext != pToBeDeleted)
{
pNode = pNode->m_pNext;
}
pNode->m_pNext = NULL;
delete pToBeDeleted;
pToBeDeleted = NULL;
}
}
參考書籍:《劍指Offer》