题解 | #字符串变形#

class Solution {
public:
    string trans(string s, int n) {
        // write code here
        for (int i = 0; i < n; i++) {
            if (s[i] >= 'a' && s[i] <= 'z')
                s[i] = toupper(s[i]);
            else 
                s[i] = tolower(s[i]);
        }

        reverse(s.begin(), s.end());
        for (int i = 0, j = 0; i < n; i++) {
            if (s[i] != ' ') {
                j = i;
                while (isalnum(s[j])) j++;
                reverse(s.begin() + i, s.begin() + j);
                i = j;
            }
        }

        return s;
    }
};

• 思路：大小写转换+翻转单词
• 1、先转换大小写
• 2、翻转整个字符串，然后翻转每个单词
• 时间复杂度：O(n)
• 空间复杂度：O(n)