05.替换空格
过于简单
- 核心思想在于新建一个string来方便操作
- 使用insert来插入字符串或者字符
字符串的插入和删除
class Solution {public:string replaceSpace(string s) {string ret;for(int i = 0; i < s.size(); i++){if(s[i] == ' '){ret.insert(ret.size(), "%20");continue;}ret.insert(ret.size(), 1, s[i]);}return ret;}};
58.-Ⅱ左旋转字符串
这一题虽然简单但是有牛比写法先贴出自己的笨比写法
class Solution {public:string reverseLeftWords(string s, int n) {string str = s.substr(0, n);//这里的0表示从0开始，n代表多少位s.erase(0, n);s.insert(s.size(), str);//这里的在指定位置插入字符串不包括插入前的最后一位return s;}};
究极靓仔写法
直接复制一段拼接在后面，然后取子串就行了 ```cpp class Solution { public: string reverseLeftWords(string s, int n) {
} };int len = s.size();s += s;return s.substr(n,len);
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### [剑指 Offer 20. 表示数值的字符串](https://leetcode-cn.com/problems/biao-shi-shu-zhi-de-zi-fu-chuan-lcof/)
基本就是暴力解法了
```cpp
class Solution {
public:
enum State {
STATE_INITIAL,
STATE_INT_SIGN,
STATE_INTEGER,
STATE_POINT,
STATE_POINT_WITHOUT_INT,
STATE_FRACTION,
STATE_EXP,
STATE_EXP_SIGN,
STATE_EXP_NUMBER,
STATE_END
};
enum CharType {
CHAR_NUMBER,
CHAR_EXP,
CHAR_POINT,
CHAR_SIGN,
CHAR_SPACE,
CHAR_ILLEGAL
};
CharType toCharType(char ch) {
if (ch >= '0' && ch <= '9') {
return CHAR_NUMBER;
} else if (ch == 'e' || ch == 'E') {
return CHAR_EXP;
} else if (ch == '.') {
return CHAR_POINT;
} else if (ch == '+' || ch == '-') {
return CHAR_SIGN;
} else if (ch == ' ') {
return CHAR_SPACE;
} else {
return CHAR_ILLEGAL;
}
}
bool isNumber(string s) {
unordered_map<State, unordered_map<CharType, State>> transfer{
{
STATE_INITIAL, {
{CHAR_SPACE, STATE_INITIAL},
{CHAR_NUMBER, STATE_INTEGER},
{CHAR_POINT, STATE_POINT_WITHOUT_INT},
{CHAR_SIGN, STATE_INT_SIGN}
}
}, {
STATE_INT_SIGN, {
{CHAR_NUMBER, STATE_INTEGER},
{CHAR_POINT, STATE_POINT_WITHOUT_INT}
}
}, {
STATE_INTEGER, {
{CHAR_NUMBER, STATE_INTEGER},
{CHAR_EXP, STATE_EXP},
{CHAR_POINT, STATE_POINT},
{CHAR_SPACE, STATE_END}
}
}, {
STATE_POINT, {
{CHAR_NUMBER, STATE_FRACTION},
{CHAR_EXP, STATE_EXP},
{CHAR_SPACE, STATE_END}
}
}, {
STATE_POINT_WITHOUT_INT, {
{CHAR_NUMBER, STATE_FRACTION}
}
}, {
STATE_FRACTION,
{
{CHAR_NUMBER, STATE_FRACTION},
{CHAR_EXP, STATE_EXP},
{CHAR_SPACE, STATE_END}
}
}, {
STATE_EXP,
{
{CHAR_NUMBER, STATE_EXP_NUMBER},
{CHAR_SIGN, STATE_EXP_SIGN}
}
}, {
STATE_EXP_SIGN, {
{CHAR_NUMBER, STATE_EXP_NUMBER}
}
}, {
STATE_EXP_NUMBER, {
{CHAR_NUMBER, STATE_EXP_NUMBER},
{CHAR_SPACE, STATE_END}
}
}, {
STATE_END, {
{CHAR_SPACE, STATE_END}
}
}
};
int len = s.length();
State st = STATE_INITIAL;
for (int i = 0; i < len; i++) {
CharType typ = toCharType(s[i]);
if (transfer[st].find(typ) == transfer[st].end()) {
return false;
} else {
st = transfer[st][typ];
}
}
return st == STATE_INTEGER || st == STATE_POINT || st == STATE_FRACTION || st == STATE_EXP_NUMBER || st == STATE_END;
}
};
简单一点
class Solution {
public:
bool hasE, hasNum, isFloat;
bool isNumber(string& s) {
for(int i = 0; i < s.size(); i++) {
auto c = s[i];
// 符号位只能在首位或E的后一位
if((c == '-' || c == '+') && (i == 0 || s[i - 1] == 'e' || s[i - 1] == 'E')) {
}else if((c == 'e' || c == 'E') && !hasE && hasNum) {
//只存在一个E, 前面必须有数字, 后面也必须有数字
hasE = true;
hasNum = false;
}else if(c == '.' && !isFloat && !hasE) {
//只存在一个小数点, 不能在E的后面
isFloat = true;
}else if(isdigit(c)) {
hasNum = true;
}else {
return false;
}
}
return hasNum;
}
};
剑指 Offer 67. 把字符串转换成整数
一般都是使用ascll码来写的
class Solution {
private:
int to_int(string& s) {
int num = 0;
int len = s.length() - 1;
bool neg = false;
for(int i = 0; i <= len; i++) {
if(i == 0 && s.at(i) == '-') {
//如果有负号，标记一下
neg = true;
continue;
}
//#define INT_MAX 2147483647
//这里小心，可能是2147483645 这样的数是不会越界所以需要特判
if((num == INT_MAX/10 && s.at(i)>'7') || num > INT_MAX/10) {
if(neg) return INT_MIN;
return INT_MAX;
}
num = num * 10 + (s.at(i)-'0');
}
if(neg) num = -num;
return num;
}
bool is_num(char& c) {
return c >= '0' && c <= '9';
}
public:
int strToInt(string str) {
string num;
int len = str.length() - 1;
int i = 0;
//去除前缀' '
while(i <= len && str.at(i) == ' ') i++;
//空串返回0
if(i > len) return 0;
//获取非空首个字符
char start = str.at(i);
//如果只剩一个字符且不为数字，返回0
if(!is_num(start) && i == len) return 0;
//符合题意的情况
//1.开头是符号且跟着数字的，2.开头就是数字的情况
else if(is_num(str.at(i)) || (( start == '-' || start == '+' ) && i < len && is_num(str.at(++i))) ) {
//'+'号直接舍去
if(start == '-') {
num += move(start);
}
//num拼接后序所有数字字符
while(i <= len && is_num(str.at(i))) {
num += move(str.at(i));
i++;
}
}
return to_int(num);
}
};
。
剑指 Offer II 032. 有效的变位词（数组代表字符串）
用一个26位的数组来表示字符串
class Solution { public: bool isAnagram(string s, string t) { if (s.length() != t.length() || s == t) { return false; } vector<int> table(26, 0); for (auto& ch: s) { table[ch - 'a']++; } for (auto& ch: t) { table[ch - 'a']--; if (table[ch - 'a'] < 0) { return false; } } return true; } };剑指 Offer II 035. 最小时间差（排序）
东西不多，先排序然后相邻的进行比较
- 唯一值得在意的点就是如果输入的个数大于1440就证明一定有重复的，直接返回0即可。
```cpp
class Solution {
//转换成分钟
int getMinutes(string &t) {
}return (int(t[0] - '0') * 10 + int(t[1] - '0')) * 60 + int(t[3] - '0') * 10 + int(t[4] - '0');
public:
int findMinDifference(vector
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