字符串的相关操作

344. 反转字符串

方法：模拟

class Solution {
public:
    void reverseString(vector<char>& s) {
        int n = s.size(), len = n / 2;
        for (int i = 0; i < len; ++i) {
            swap(s[i], s[n-i-1]);
        }
    }
};

$时间复杂度O(n), 空间复杂度O(1);

541. 反转字符串 II

方法：双指针

class Solution {
public:
    string reverseStr(string s, int k) {
        int n = s.size(), start = 0, end;
        int i, j;
        for (i = 0; start < n; i += 2*k) {
            end = start + 2 * k;
            if (end > n) end = n;
            if (end - start >= k && end - start <= 2 * k) {
                for (j = start; j < start + k / 2; ++j) swap(s[j], s[2*start+k-j-1]);
            } else {
                for (j = start; j < start + (end-start)/2; ++j) swap(s[j], s[end-j-1+start]);
            }
            start = end;
        }
        return s;
    }
};

$时间复杂度O(n*k), 空间复杂度O(1);

剑指 Offer 05. 替换空格

方法：模拟

class Solution {
public:
    string replaceSpace(string s) {
        string str;
        for (char& c : s) {
            if (c == ' ') {
                str += "%20";
            } else str += c;
        }
        return str;
    }
};

$时间复杂度O(n), 空间复杂度O(1);

151. 反转字符串中的单词

方法：模拟

class Solution {
private:
    void removeExtraspace(string& s) {
        int slow = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] != ' ') {
                if (slow != 0) s[slow++] = ' ';
                while (s[i] != ' ' && i < s.size()) s[slow++] = s[i++];
            }
        }
        s.resize(slow);
    }
public:
    string reverseWords(string s) {
        removeExtraspace(s);
        reverse(s.begin(), s.end());
        for (int i = 0, st = 0; i <= s.size(); ++i) {
            if (s[i] == ' ' || i == s.size()) {
                reverse(s.begin() + st, s.begin() + i);
                st = i + 1;
            }
        }
        return s;
    }
};

$时间复杂度O(), 空间复杂度O(n);

剑指 Offer 58 - II. 左旋转字符串

方法：模拟

class Solution {
public:
    string reverseLeftWords(string s, int k) {
        reverse(s.begin(), s.begin()+k);
        reverse(s.begin()+k, s.end());
        reverse(s.begin(), s.end());
        return s;
    }
};

$时间复杂度O(n), 空间复杂度O(1);