常用数据结构之字符串

Posted 2020-09-10 悠悠南山下

针对字符串处理中一些经常遇到的问题进行总结

/*
     * 字符串比较问题
     * 对比两个字符串是否一致，可以使用哈希表，哈希之后，对比哈希表是否一致即可
     */
    bool chkTransform(string A, int lena, string B, int lenb) {
        int *hash = new int[256];
        memset(hash, 0, sizeof(int) * 256);
        for (int i = 0; i < lenb; i++) {
            hash[A[i]]++;
        }
        for (int i = 0; i < lenb; i++) {
            if (hash[B[i]]-- == 0)
                return false;
        }
        return true;
    }
    /*
     * 旋转词问题
     * 如果B的元素可以全部在A中找到，则B是A的旋转词
     * 解法：A+A穷举了A所有可能的旋转词，用B在A+A中做匹配即可
     */
    bool chkRotation(string A, int lena, string B, int lenb) {
        if (A.empty() || B.empty() || lena != lenb)
            return false;
        string C = A + A;
        string D = B + B;
        if ((C.find(B.c_str(), 0, lenb) != string::npos)
                && (D.find(A.c_str(), 0, lena) != string::npos))
            return true;
        return false;
    }
    /*
     * 翻转句子，以空格为分隔符
     */
    static void reverse(string &A, int begin, int end) {
        int mid = (begin + end) / 2;
        for (int i = begin; i <= mid; i++) {
            swap(A[i], A[end - i + begin]);
        }
    }
    static string reverseSentence(string A, int n) {
        if (n <= 1)
            return A;
        string res(A);
        reverse(res, 0, n - 1);
        int begin = 0;
        int end = 0;
        for (int i = 0; i < n; i++) {
            if (res[i] == ‘ ‘) {
                end = i - 1;
                reverse(res, begin, end);
                begin = i + 1;
            }
        }
        reverse(res, begin, n - 1);
        return res;
    }
    /*
     * 给出一些短字符串，拼接得到按字母表最小的字符串
     */
    static bool compare(string a, string b) {
        string ab = a + b;
        string ba = b + a;
        return ab < ba ? true : false;
    }
    string findSmallest(vector<string> strs, int n) {
        string res;
        if (n == 0)
            return res;
        if (n == 1)
            return strs[0];
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                if (compare(strs[j], strs[i])) {
                    swap(strs[i], strs[j]);
                }
            }
            res += strs[i];
        }
        return res;
    }
    /*
     * 循环移位
     */
    static string stringTranslation(string A, int n, int len) {
        if (n <= 1)
            return A;
        string res(A);
        int l = len % n;
        reverse(res, 0, l - 1);
        reverse(res, l, n - 1);
        reverse(res, 0, n - 1);
        return res;
    }
    /*
     * 替换字符串中的空格
     */
    string replaceSpace(string iniString, int length) {
        if (length < 1)
            return iniString;
        int count = 0;
        for (int i = 0; i < length; i++) {
            if (iniString[i] == ‘ ‘)
                count++;
        }
        int newLength = 2 * count + length;
        int pos = newLength - 1;
        iniString.append(2 * count, ‘ ‘);
        for (int i = length - 1; i >= 0; i--) {
            if (iniString[i] == ‘ ‘) {
                iniString[pos--] = ‘0‘;
                iniString[pos--] = ‘2‘;
                iniString[pos--] = ‘%‘;
            } else {
                iniString[pos--] = iniString[i];
            }
        }
        return iniString;
    }
    /*
     * 判断字符串中左右括号是否配对
     */
    bool chkParenthesis(string A, int n) {
        if (A.empty() || n == 0)
            return false;
        int num = 0;
        for (int i = 0; i < n; i++) {
            if (A[i] != ‘(‘ && A[i] != ‘)‘) {
                return false;
            }
            if (A[i] == ‘)‘) {
                num--;
                if (num < 0)
                    return false;
            }
            if (A[i] == ‘(‘) {
                num++;
            }
        }
        return num == 0;
    }
    /*
     * 给定一个字符串，求最大无重复字符子串的长度
     */
    int longestSubstring(string A, int n) {
        int last[256];
        for (int i = 0; i < 256; ++i)
            last[i] = -1;

        last[A[0]] = 0;
        int pre = 1;
        int max = 1;

        for (int i = 1; i < n; ++i) {
            if (last[A[i]] < i - pre) {
                ++pre;
            } else {
                pre = i - last[A[i]];
            }
            last[A[i]] = i;
            if (max < pre)
                max = pre;
        }
        return max;
    }
    /*
     * 将字符串转换成整数
     */
    static int StrToInt(string str) {
        if (str.empty())
            return 0;
        int space = 0;
        while (str[space] == ‘ ‘) {
            space++;
        }
        int flag = 1;
        str = str.substr(space, str.length() - space + 1);
        vector<int> v;
        for (int i = 0; i < str.length(); i++) {
            if (i == 0) {
                if (str[0] == ‘-‘) {
                    flag = 0;
                } else if (str[0] == ‘+‘) {
                    flag = 1;
                } else if (isdigit(str[0])) {
                    flag = 1;
                    v.push_back(str[0] - ‘0‘);
                } else {
                    return 0;
                }
            } else {
                if (!isdigit(str[i]))
                    return 0;
                else
                    v.push_back(str[i] - ‘0‘);
            }
        }
        if (v.empty())
            return 0;
        long long num = v[0];
        for (int i = 1; i < v.size(); i++) {
            num = num * 10 + v[i];
        }
        if (flag == 0)
            num = -num;
        return num;
    }
    /*
     * 判断回文
     * 从两端到中间
     */
    bool isPalindrome(char *str) {
        if (str == NULL)
            return false;
        int length = strlen(str);
        //cout << "length= " << length << endl;
        int head = 0;
        int tail = length - 1;
        while (head <= tail) {
            if (str[head] == str[tail]) {
                head++;
                tail--;
            } else {
                return false;
            }
        }
        return true;
    }
    /*
     * 先从中间开始、然后向两边扩展查看字符是否相等
     *
     */
    bool isPalindrome2(char *str) {
        if (str == NULL)
            return false;
        int length = strlen(str);
        int mid;
        int tohead;
        int totail;
        if (length % 2 == 0) {
            mid = length / 2;
            tohead = mid - 1;
            totail = mid;
        } else {
            mid = (length - 1) / 2;
            tohead = mid - 1;
            totail = mid + 1;
        }
        while (tohead >= 0) {
            if (str[tohead--] != str[totail++])
                return false;
        }
        return true;
    }
    /*
     * 最长回文子串
     */
    int longestPalindrom(const char *str) {
        if (str == NULL)
            return 0;
        int length = strlen(str);
        int max = 0;
        int c = 0;
        for (int i = 0; i < length; i++) {
            for (int j = 0; (i - j >= 0) && (i + j) < length; j++) {
                if (str[i - j] != str[i + j])
                    break;
                c = 2 * j + 1;
            }
            if (c > max)
                max = c;
            for (int j = 0; (i - j >= 0) && (i + j) < length; j++) {
                if (str[i - j] != str[i + j + 1])
                    break;
                c = 2 * j + 2;
            }
            if (c > max)
                max = c;
        }
        return max;
    }
    /*
     * 字符串全排列
     */
    void cal(vector<string>&v, int k, string str) {
        if (k == str.size() - 1)
            v.push_back(str);
        set<char> s;
        sort(str.begin() + k, str.end());
        for (int i = k; i < str.size(); i++) {
            if (s.find(str[i]) == s.end()) {
                s.insert(str[i]);
                swap(str[i], str[k]);
                cal(v, k + 1, str);
                swap(str[i], str[k]);
            }
        }
    }
    vector<string> Permutation(string str) {
        vector < string > v;
        cal(v, 0, str);
        return v;
    }
    /*
     * 数组中连续子序列和最大值
     */
    int maxSubSequence(vector<int> arr) {
        int maxSum = arr[0];
        int curSum = 0;
        for (int i = 0; i < arr.size(); i++) {
            curSum = (arr[i] > curSum + arr[i]) ? arr[i] : curSum + arr[i];
            maxSum = (curSum > maxSum) ? curSum : maxSum;
        }
        return maxSum;
    }
    /*
     * longest common prefix
     */
    static string commonPrefix(vector<string>&strs) {
        string res;
        if (strs.size() < 1)
            return res;
        char prefix; //不要直接赋‘a’，因为你也不知道第一个是什么
        bool isprefix = true;
        for (int j = 0; j < strs[0].size(); j++) {
            prefix = strs[0][j];
            for (int i = 0; i < strs.size(); i++) {
                if (strs[i][j] != prefix) {
                    isprefix = false;
                    break;
                }
            }
            if (!isprefix)
                break;
            res += prefix;
        }
        return res;
    }