71. 简化路径

自己做

解：遍历检查

class Solution {
public:string simplifyPath(string path) {int p = 0;string res;while(p < (int)path.size()){//情况1：遇到"/./" =》p跳过"/."if(p < (int)path.size() - 2 && path[p] == '/' && path[p + 1] == '.' && path[p + 2] == '/')p += 2;//情况2：遇到"//" =》p跳过"/"else if(p < (int)path.size() - 1 && path[p] == '/' && path[p + 1] == '/')p++;        //情况3_1：遇到"/abc" =》正常输入else if(p < (int)path.size() - 1 && path[p] == '/' && path[p + 1] != '/' && path[p + 1] != '.'){res.push_back(path[p]);             //加入斜杠p++;while(p < (int)path.size() && path[p] != '/'){      //加入abc ，直到遍历结束或者遇到下个斜杠res.push_back(path[p]);            p++;                    }}//情况3_2：遇到"/..xxxxx" =》正常输入else if(p < (int)path.size() - 3 && path[p] == '/' && path[p + 1] == '.' && path[p + 2] == '.' && path[p + 3] != '/'){res.push_back(path[p]);             //加入斜杠p++;while(p < (int)path.size() && path[p] != '/'){      //加入.......... ，直到遍历结束或者遇到下个斜杠res.push_back(path[p]);             p++;                    }}//情况3_3：遇到"/.xxxx" =》正常输入else if(p < (int)path.size() - 2 && path[p] == '/' && path[p + 1] == '.' && path[p + 2] != '.' && path[p + 2] != '/'){res.push_back(path[p]);             //加入斜杠p++;while(p < (int)path.size() && path[p] != '/'){      //加入.......... ，直到遍历结束或者遇到下个斜杠res.push_back(path[p]);             p++;                    }}//情况4：遇到"/../"或者"/.."(末尾) =》弹出上级目录，p跳过"/.."【在检查完有没有/...的情况后】else if(p < (int)path.size() - 2 && path[p] == '/' && path[p + 1] == '.' && path[p + 2] == '.'){p += 3;while(res.size() > 0 && res[(int)res.size() - 1] != '/')   //对于res = "xxx/abc"，弹出abc这个上级目录，得到结果res = "xxx/"res.pop_back();if(res.size() >= 1)     //弹出斜杠res.pop_back();} //情况5：末尾不成目录"abc/" =》 结束遍历elsebreak;}//防止空目录if(res.size() == 0)res.push_back('/');return res;}
};

今日总结