leetcode 字符串类型题

1，Vaild Palindrome

bool isPalindrome(string& s) {
    transform(s.begin(), s.end(), s.begin(), tolower); // 把字符全部转换成小写
    int left = 0;
    int right = s.length()-1;
    while (left < right) {
        if (!isalnum(s[left])) ++left;
        else if (!isalnum(s[right])) --right;
        else if (s[left] != s[right]) return false;
        else {
            ++left;
            --right;
        }
    }
    return true;
}

2，Implement strStr

int strStr(const string& haystack, const string& needle) { // KMP 算法，BM 算法
    if (needle.empty()) return 0;
    const int N = haystack.length() - needle.length();
    for (int i = 0; i <= N; ++i) {
        int j = i;  // 源串索引
        int k = 0;  // 字串索引
        while (haystack[j] == needle[k] &&  j < haystack.length() && k < needle.length()) {
            ++j;
            ++k;
        }
        if (k == needle.length()) return i;
    }
    return -1;
}

3，String to Integer(atoi)

int myAtoi(const string& str) {
    const int n = str.length();
    int sign = 1;
    int i = 0;
    int num = 0;

    while (str[i] == ' ' && i < n) ++i;  // 前面的字符处理
    if (str[i] == '+') {
        sign = 1;
        ++i;
    }
    else if (str[i] == '-') {
        sign = -1;
        ++i;
    }
    for (; i < n; ++i) {
        if (str[i] < '0' || str[i] > '9')
            break;
        num = num * 10 + str[i] - '0';

        if (sign == 1 && num > INT_MAX)
            return INT_MAX;
        else if (sign == -1 && num > INT_MIN + 1)
            return INT_MIN;
    }
    return sign * num;
}

4，Add Binary

string addBinary(string a, string b) {  // 思路同 Add Binary(链表类型题) plusOne(数组类型题)
    string result;
    int carry = 0;
    reverse(a.begin(), a.end()); // 先反转两个数组，从低位开始相加
    reverse(b.begin(), b.end());
    const int n = a.length() > b.length() ? a.length() : b.length();
    for (int i = 0; i < n; ++i) {
        const int ai = i < a.length() ? a[i]-'0' : 0;
        const int bi = i < b.length() ? b[i]-'0' : 0;
        int value = (ai + bi + carry) % 2;
        carry = (ai + bi + carry) / 2;
        result.insert(result.begin(), value + '0');
    }
    if (carry == 1)
        result.insert(result.begin(), '1');
    return result;
}

5，Longest Palindromic Substring(动态规划)

int handler(string& s) {
    int len = s.length();
    int maxLen = INT_MIN;
    //dp[i][j] 表示 s[i:j] 的字符串是否是回文字符串
    vector<vector<bool>> dp(len, vector<bool>(len, false));
    if(len == 0 || len == 1) return len;

    for(int j=0; j<len; j++) {  // 竖着初始化，初始化矩阵的上三角
        for(int i=0; i<=j; i++) {
            if(j - i < 2)
                dp[i][j] = s[i] == s[j];
            else
                //判断 dp[i][j] 的左下角的元素 dp[i+1][j-1]
                dp[i][j] = dp[i+1][j-1] && s[i] == s[j];
            //取最大长度
            if(dp[i][j] && j-i+1 > maxLen)
                maxLen = max(maxLen, j-i+1);
        }
    }
    return maxLen;
}

6，Regular Expression Matching

bool isMatchI(const char *s, const char *p) {   //递归版   有挑战的一道题目
    if (*p == '\0') return *s == '\0';

    //next char is not '*',then match current character
    if (*(p + 1) != '*') {
        if (*p == *s || (*p == '.' && *s != '\0')) { // correctly match
            return isMatchI(s + 1, p + 1);
        }
        else {                                      // failed match
            return false;
        }
    }
    else {                                           // next char is '*'
        while (*p == *s || (*p == '.' && *s != '\0')) {
            if (isMatchI(s, p + 2)) {
                return true;
            }
            s++;
        }
        return isMatchI(s, p + 2);
    }
}

7，Wildcard Matching

bool isMatchII1(const char *s, const char *p) {  // 递归版
    if (*p == '*') {
        while (*p == '*') ++p;  // skip continuous '*'
        if (*p == '\0') return true;
        while (*s != '\0' && !isMatchII1(s, p)) ++s;

        return *s != '\0';
    }
    else if (*p == '\0' || *s == '\0')
        return *p == *s;
    else if (*p == *s || *p == '?')
        return isMatchII1(++s, ++p);
    else
        return false;
}
bool isMatchII2(const char *s, const char *p) {  // 迭代版
    bool star = false;
    const char *str = s;  // str 可以变，*str 不能变
    const char *ptr = p;

    while (*str != '\0') {
        switch (*ptr) {
        case '?':
            str++;
            ptr++;
            break;
        case '*':
            star = true;
            while (*ptr == '*') ++p;
            if (*ptr == '\0') return true;
            break;
        default:
            if (*str != *ptr) {
                if (!star) 
                    return false;
                str++;
            }
        }
    }
    while (*ptr == '*') ++ptr;
    return (*ptr == '\0');

}

8，Longest Common Prefix

string longestCommonPrefix1(vector<string>& strs) {  // 纵向扫描
    if (strs.empty()) return "";

    for (int index = 0; index < strs[0].size(); ++index) {  // 选取第一个字符串和其它字符串进行比较
        for (int i = 1; i < strs.size(); ++i) {
            if (strs[i][index] != strs[0][index])
                return strs[0].substr(0, index);
        }
    }
    return strs[0];
}

string longestCommonPrefix2(vector<string>& strs) { // 横向扫描
    if (strs.empty()) return "";

    int right_most = strs[0].size() - 1;
    for (size_t i = 1; i < strs.size(); ++i) {
        for (int j = 0; j <= right_most; ++j) {
            if (strs[i][j] != strs[0][j])
                right_most = j - 1;
        }
    }
    return strs[0].substr(0, right_most + 1);
}

9，Valid Number

bool isNumber(const char *s) {  // 
    char* endptr;
    strtod(s, &endptr);
    
    if (endptr == s)
        return false;

    for (; *endptr; ++endptr) {
        if (!isspace(*endptr))
            return false;
    }
    return true;
}

10，Integet to Roman

string intToRoman1(int num) {
    const int radix[] = { 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };
    const string symbol[] = { "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" };
    
    string roman;
    for (size_t i = 0; num > 0; ++i) {
        int count = num / radix[i];
        num %= radix[i];
        for (; count > 0; --count)
            roman += symbol[i];
    }
    return roman;
}

string intToRoman2(int num) {
    string res = "";
    vector<int> val = { 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };
    vector<string> str = { "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" };
    for (int i = 0; i < val.size(); ++i) {
        while (num >= val[i]) {
            num -= val[i];
            res += str[i];
        }
    }
    return res;
}

11，Roman to Integer

int romanToInt(const string& s) {
    int result = 0;
    unordered_map<char, int> mapping;
    mapping['I'] = 1;
    mapping['V'] = 5;
    mapping['X'] = 10;
    mapping['L'] = 50;
    mapping['C'] = 100;
    mapping['D'] = 500;
    mapping['M'] = 1000;

    for (size_t i = 0; i < s.size(); i++) {
        if (i > 0 && (mapping[i] > mapping[i - 1])) {
            result += (mapping[s[i]] - 2 * mapping[s[i - 1]]);
        }
        else {
            result += mapping[s[i]];
        }
    }
    return result;
}

12，Count and Say

string getNext(const string& s) {
    stringstream ss;
    for (auto i = s.begin(); i != s.end();) {
        auto j = find_if(i, s.end(), bind1st(not_equal_to<char>(), *i));  // 需要看看函数怎么用
        ss << distance(i, j) << *i;
        i = j;
    }
    return ss.str();
}

string countAndSay(int n) {
    string s("1");
    while (--n) {
        s = getNext(s);
    }
    return s;
}

13，Anagrams(回文构词法)

vector<string> anagrams(vector<string>& strs) {
    unordered_map<string, vector<string>> group;
    for ( auto s = strs.begin(); s != strs.end();++s) {
        string key = *s;
        sort(key.begin(), key.end());  // 会修改原数据
        group[key].push_back(*s);
    }
    vector<string> result;
    for (auto it = group.cbegin(); it != group.cend(); ++it) {
        if (it->second.size() > 1)
            result.insert(result.end(), it->second.begin(), it->second.end());
    }
    return result;
}

14，Simplify Path

string simplifyPath(const string& path) {
    string dir;  
    stack<string> stk;
    string result;

    for (auto i = path.begin(); i != path.end();) {
        ++i;
        auto j = find(i, path.end(), '/');
        dir = string(i, j);   /*  获取两个 / / 之间的内容 */

        if (!dir.empty() && dir !="/" && dir != ".") {
            if (dir == "..") {
                if (!stk.empty())
                    stk.pop();
            }
            else
                stk.push(dir);
        }
        i = j;
    }
    //stringstream out;  // 可以用字符串流保存，然后通过 out.str() 转化成字符串返回
    if (stk.empty())
        //out << "/";
        result = "/";
    else {
        while (!stk.empty()) {
            string s = stk.top();
            stk.pop();
            //out << '/' << s;
            result += "/" + s;
        }
    }
    // return out.str();
    return result;
}

15，Length of Last Word

int lengthOfLastWord1(const string& s) {  // STL::find_if,find_if_not,distance
    auto first = find_if(s.rbegin(), s.rend(), isalpha);
    auto last = find_if_not(first, s.rend(), isalpha);
    return distance(first, last);
}

int lengthOfLastWord2(const string& s) {
    int len = 0;
    for (int i = 0; i < s.length(); ++i) {
        if (s[i] != ' ')
            ++len;
        else
            len = 0;
    }
    return len;
}

 

c++常用函数

1, split

void split(string& s, vector<string>& v, char deli) {
    int current = 0, next = -1;
    do {
        current = next + 1;
        next = s.find_first_of(deli, current);
        v.push_back(s.substr(current, next - current));
    }while(next != -1);
}

 

以上题目来源于：http://www.github.com/soulmachine/leetcode