题解 | #字符串变形#

class Solution {
public:
  //用于字母大小写转换
    void change(string& s, int n) {
        for (int i = 0; i < n; i++) {
            if (s[i] >= 'a' && s[i] <= 'z') {
                s[i] -= 32;
            }
            else if (s[i] >= 'A' && s[i] <= 'Z') {
                s[i] += 32;
            }
            else {
                continue;
            }
        }
    }
  //用于反转字符串
    void reserve(string& s, int left,int right) {
        char ch;
        while (left < right)
        {
            ch = s[left];
            s[left] = s[right];
            s[right] = ch;
            left++;
            right--;
        }
    }
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param s string字符串 
     * @param n int整型 
     * @return string字符串
     */

    string trans(string s, int n) {
        if (s == "" || n <= 0) return "";
	  //1、转换字母大小写
        change(s, n);
	  //2、反转整个字符串
        reserve(s,0, n-1);
	  //3、反转每个单词，当快指针走到'\0'处结束
	    int slow = 0;
        int fast = 0;
        while (1) {
            if (s[fast] == ' '||s[fast]=='\0') {
                reserve(s, slow, fast-1);
                if (fast == n) break;
                fast++;
                slow = fast;
            }
            else {
                fast++;
            }
        }
        return s;
    }
    //比如：Hello World
    //第一步处理大小写：hELLO wORLD
    //第二步反转整个字符串：DLROw OLLEh
    //第三步反转每个单词：wORLD hELLO
};