LeetCode字符串

LeetCode字符串

LeetCode.38 外观数列

class Solution {
    
public:
    string countAndSay(int n) {
    
        string s = "1";
        for(int i = 1; i < n; ++i) {
    
            string ns;
            for(int j = 0; j < s.size(); ++j) {
    
                int k = j;
                while(k < s.size() && s[k] == s[j]) ++k;
                ns += to_string(k-j) + s[j];
                j = k-1;    // 因为for循环结束一次j要+1，所以这里-1;
            }
            s = ns;
        }
        return s;
    }
};

LeetCode.49 字母异位词分组

class Solution {
    
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
    
        unordered_map<string,vector<string>> hash;
        for(auto str:strs) {
    
            string t = str;
            sort(t.begin(), t.end());
            hash[t].push_back(str);
        }
        vector<vector<string>> ans;
        for(auto h : hash) ans.push_back(h.second);
        return ans;
    }
};

LeetCode.151 颠倒字符串中的单词

class Solution {
    
public:
    string reverseWords(string s) {
    
        int k = 0;
        for(int i = 0; i < s.size(); ++i) {
    
            while(i < s.size() && s[i] == ' ') ++i;
            if(i == s.size()) break;
            int j = i;
            while(j < s.size() && s[j] != ' ') ++j;
            reverse(s.begin() + i, s.begin() + j);
            if(k) s[k++] = ' ';
            while(i < j) s[k++] = s[i++];
        }
        s.erase(s.begin() + k, s.end());
        reverse(s.begin(), s.end());
        return s;
    }
};

LeetCode.165 比较版本号

class Solution {
    
public:
    int compareVersion(string version1, string version2) {
    
        int i = 0, j = 0;
        // 两个字符串都遍历完再结束循环
        while(i < version1.size() || j < version2.size()) {
    
            int x = i, y = j;
            while(x < version1.size() && version1[x] != '.') ++x;
            while(y < version2.size() && version2[y] != '.') ++y;
            int a = x == i? 0 : atoi(version1.substr(i, x-i).c_str());// 两个字符串长度可能不一致，如果有一个字符串先遍历完，0
            int b = y == j? 0 : atoi(version2.substr(j, y-j).c_str());
            if(a > b) return 1;
            if(a < b) return -1;
            i = x+1, j = y+1;
        }
        return 0;
    }
};

LeetCode.929 独特的电子邮件地址

class Solution {
    
public:
    int numUniqueEmails(vector<string>& emails) {
    
        unordered_set<string> hash;
        for(auto email:emails) {
    
            int index = email.find('@');
            string name;
            // 过滤
            for(auto c:email.substr(0, index)) {
    
                if(c == '+') break;
                else if(c != '.') name += c;
            }
            string domain = email.substr(index);
            hash.insert(name + domain);
            // cout << name + domain << endl;
        }
        return hash.size();
    }
};

LeetCode.5 最长回文子串

class Solution {
    
public:
    string longestPalindrome(string s) {
    
        string ans;
        for(int i = 0; i < s.size(); ++i) {
    
            for(int j = i, k = i; j >=0 && k < s.size() && s[j]==s[k]; --j,++k)
                if(k-j+1 > ans.size())
                    ans = s.substr(j, k-j+1);
        }
        for(int i = 0; i < s.size(); ++i) {
    
            for(int j = i, k = i+1; j >=0 && k < s.size() && s[j]==s[k]; --j,++k)
                if(k-j+1 > ans.size())
                    ans = s.substr(j, k-j+1);
        }
        return ans;
    }
};

LeetCode.6 Z字形变换

找规律：

• 第0行是首项为0，公差为2*(n - 1)的等差数列；
• 中间的是两个等差数列交错，每相邻的两个数为一组，和为2*(n - 1)；
• 第n - 1行是首项为n - 1，公差为2*(n - 1)的等差数列；

class Solution {
    
public:
    string convert(string s, int numRows) {
    
        // 处理numRows = 1时的情况
        if(numRows == 1) return s;

        string ans;
        for(int i = 0; i < numRows; ++i) {
    
            if(!i || i == numRows - 1) {
    
                for(int j = i; j < s.size(); j += 2 * (numRows - 1))
                    ans += s[j];
            }
            else {
    
                for(int j = i, k = 2*(numRows-1)-i; j < s.size() || k < s.size();j += 2*(numRows-1), k += 2*(numRows-1))     {
    
                    if(j < s.size()) ans += s[j];
                    if(k < s.size()) ans += s[k];
                }   
            }
        }
        return ans;
    }
};

LeetCode.3 无重复字符的最长子串

class Solution {
    
public:
    int lengthOfLongestSubstring(string s) {
    
        unordered_map<char, int> hash;
        int res = 0;
        for(int i = 0, j = 0; i < s.size(); ++i) {
    
            hash[s[i]]++;
            // 每次循环往不重复子串中新添加的可能是不重复的，可能是重复的
            // 如果发现新添加了之后是重复的，就移动左边界j，
            while(hash[s[i]] > 1) hash[s[j++]] --;
            res = max(res, i - j + 1);
        }
        return res;
    }
};

LeetCode.273 整数数组转换英文表示

分块处理：

— Billion — Million — Thousand —

(1-999) (1-999) (1-999) (1-999)

class Solution {
    
public:
    string small[20] = {
    "Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
    string decade[10] = {
    "", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
    string big[4] = {
    "Billion", "Million", "Thousand", ""};
    string numberToWords(int num) {
    
        string res;
        if(!num) return small[num];
        for(int i = 1000000000, j = 0; i > 0; i/=1000, ++j) {
    
            if(num >= i) {
    
                res += get_part(num / i) + big[j] + " ";
                num %= i;
            }
        }
        while(res.back() == ' ') res.pop_back();
        return res;
    }
    string get_part(int num) {
    
        string res;
        if(num >= 100) {
    
            res += small[num / 100] + " Hundred ";
            num %= 100;
        }
        if(!num) return res;
        if(num >= 20) {
    
            res += decade[num / 10] + " ";
            num %= 10;
        }
        if(!num) return res;
        res += small[num] + " ";
        return res;
    }
};