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leetcode 71 Simplify Path ------ java
Given an absolute path for a file (Unix-style), simplify it.
For example,
path = "/home/", => "/home"
path = "/a/./b/../../c/", => "/c"
click to show corner cases.
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".
这道题就是化简地址，有两个规则： /./ 代表不变
/../ 代表返回上级目录
还有就是Corner Cases中的特殊情况都有说明
这道题刚开始写了好半天，其实主要原因是，题目中有一个没有说明，那就是带有转义字符的情况：如果目录名字中含有‘／’，如果考虑这种情况会复杂许多。但是这道题其实没有考虑这种情况，也就是说将\这个符号当成一般的字符处理。所以也就没什么难点，就是考虑清楚情况。
可以使用自动机求解。
public class Solution { public String simplifyPath(String path) { int len = path.length(); if (len == 0) return path; Stack<Character> stack = new Stack<Character>(); int i = 0; while (i < len) { if (path.charAt(i) == ‘/‘) { while( i<len && path.charAt(i) == ‘/‘) i++; i--; if( i+1 == len) i++; else if( path.charAt(i+1) == ‘.‘){ if( i+2 == len || path.charAt(i+2) == ‘/‘) i+=2; else if( path.charAt(i+2) == ‘.‘){ if( i+3 == len || path.charAt(i+3) == ‘/‘){ int j = 0; while( j<stack.size() ) if( stack.pop() == ‘/‘) break; j++; i+=3; }else{ stack.push(‘/‘); stack.push(‘.‘); stack.push(‘.‘); i+=3; while (i < len) { if (path.charAt(i) == ‘/‘) break; else stack.push(path.charAt(i)); i++; } } }else{ stack.push(‘/‘); stack.push(‘.‘); i+=2; while (i < len) { if (path.charAt(i) == ‘/‘) break; else stack.push(path.charAt(i)); i++; } } }else{ i++; stack.push(‘/‘); while (i < len) { if (path.charAt(i) == ‘/‘) break; else stack.push(path.charAt(i)); i++; } } } else { if( path.charAt(i) == ‘.‘){ if( i+1 == len || path.charAt(i+1) == ‘/‘) i+=2; else if( i+2 == len || (path.charAt(i+1) == ‘.‘ && path.charAt(i+2) == ‘/‘)) i+=3; }else{ stack.push(‘/‘); while (i < len) { if (path.charAt(i) == ‘/‘) break; else stack.push(path.charAt(i)); i++; } } stack.push(‘/‘); while (i < len) { if (path.charAt(i) == ‘/‘) break; else stack.push(path.charAt(i)); i++; } } } len = stack.size(); if( len == 0) return new String("/"); char[] result = new char[len]; for (int j = len - 1; j >= 0; j--) { result[j] = stack.pop(); } return String.valueOf(result); }}
leetcode 71 Simplify Path ------ java
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