1.字符串替換
string ChReplaceFun(string &strValue, string strOriginal, string strReplace)
{
//例如 strValue = "aabbccdd";strOriginal = "a"; strReplace = "x"; 所有的要替換的a都會變成x
if (strValue.find(strOriginal) != -1)
{
while (strValue.find(strOriginal) != -1)
{
strValue = strValue.replace(strValue.find(strOriginal), 1, strReplace);
}
}
return strValue;
}
2.測試程序運行時間
#include<windows.h>
double time = 0;
double counts = 0;
LARGE_INTEGER nFreq;
LARGE_INTEGER nBeginTime;
LARGE_INTEGER nEndTime;
QueryPerformanceFrequency(&nFreq);
QueryPerformanceCounter(&nBeginTime);//開始計時
//...測試代碼
QueryPerformanceCounter(&nEndTime);//停止計時
time = (double)(nEndTime.QuadPart - nBeginTime.QuadPart) / (double)nFreq.QuadPart;//計算程序執行時間單位爲s
cout << "運行時間:" << time * 1000 << "ms" << endl;
3.整數反轉
第一種方法:
int reverse(int x) {
int rev = 0;
while(x!=0){
int pop = x%10;
x /= 10;
if(rev>INT_MAX/10 || (rev==INT_MAX/10 && pop>INT_MAX%10))
return 0;
if(rev<INT_MIN/10 || (rev==INT_MIN/10 && pop<INT_MIN%10))
return 0;
rev = rev*10 + pop;
}
return rev;
}
第二種解法:
int reverse(int x) {
long rev = 0;
while(x!=0){
int pop = x%10;
x /= 10;
rev = rev*10 + pop;
}
return rev ==(int)rev ? (int)rev : 0;
}
4.判斷整數是否爲迴文數
bool isPalindrome(int x) {
int tmp = x;
if(x<0)
return false;
long rev = 0;
while(x>0){
int pop = x%10;
x /= 10;
rev = rev*10 + pop;
}
return (int)rev == tmp;
}
5.十進制整數的反碼
將十進制轉成二進制求出反碼,最後在十進制輸出
int bitwiseComplement(int N) {
if (N == 0)
return 1;
//1.先表示二進制 全部存入vector
vector<int> vct;
while (N>0) {
vct.push_back(N % 2);
N /= 2;
}
int result = 0;
//2.找出反碼並轉換成十進制整數
int leath = vct.size();
for (int i = 0; i <leath; i++) {
int pop = vct.back();
if (pop == 0)//原碼爲0 反碼爲一才需要加
result += pow(2, leath - i - 1);
vct.pop_back();
}
return result;
}
6.字符串反轉
char s[] = "123456";
_strrev(s); //效率是reverse函數的200倍
cout << s << endl;
string str = s;
reverse(str.begin(), str.end());
cout << str.c_str() << endl;
//字符串轉字符數組 注意str所有的字符個數不能超過數組定義最大值
char buf[50];
strcpy_s(buf, str.c_str());
7.無重複字符的最長子串
傳參字符串,返回最大子串的長度
int lengthOfLongestSubstring(string s) {
if (s.empty())//輸入空字符串 長度爲0
return 0;
int result = 1;//輸入單個字符串就不需要循環 直接返回1
for (int i = 0; i < s.length() - 1; i++) {//利用循環每個字符都組成一個串
string str;
str = s[i];
for (int j = i + 1; j < s.length(); j++) {
if (str.rfind(s[j]) == s.npos) {//符合累加條件 現有的字符串沒有重複的
str += s[j];
if (str.length() > result)
result = str.length();
}
else {//出現重複
break;
}
}
}
return result;
}