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  • [LeetCode] Simplify Path 解题报告


    Given an absolute path for a file (Unix-style), simplify it.
    For example,
    path = "/home/", => "/home"
    path = "/a/./b/../../c/", => "/c"
    Corner Cases:
    • Did you consider the case where path = "/../"?
      In this case, you should return "/".
    • Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/".
      In this case, you should ignore redundant slashes and return "/home/foo".
    » Solve this problem

    [解题思路]
    利用栈的特性,如果sub string element
    1. 等于“/”,跳过,直接开始寻找下一个element
    2. 等于“.”,什么都不需要干,直接开始寻找下一个element
    3. 等于“..”,弹出栈顶元素,寻找下一个element
    4. 等于其他,插入当前elemnt为新的栈顶,寻找下一个element

    最后,再根据栈的内容,重新拼path。这样可以避免处理连续多个“/”的问题。


    [Code]
    1:       string simplifyPath(string path) {   
    2: // Start typing your C/C++ solution below
    3: // DO NOT write int main() function
    4: vector<string> stack;
    5: assert(path[0]=='/');
    6: int i=0;
    7: while(i< path.size())
    8: {
    9: while(path[i] =='/' && i< path.size()) i++; //skip the begining '////'
    10: if(i == path.size())
    11: break;
    12: int start = i;
    13: while(path[i]!='/' && i< path.size()) i++; //decide the end boundary
    14: int end = i-1;
    15: string element = path.substr(start, end-start+1);
    16: if(element == "..")
    17: {
    18: if(stack.size() >0)
    19: stack.pop_back();
    20: }
    21: else if(element!=".")
    22: stack.push_back(element);
    23: }
    24: if(stack.size() ==0) return "/";
    25: string simpPath;
    26: for(int i =0; i<stack.size(); i++)
    27: simpPath += "/" + stack[i];
    28: return simpPath;
    29: }




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  • 原文地址:https://www.cnblogs.com/codingtmd/p/5078953.html
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