LeetCode071 Simplify Path

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详细见：leetcode.com/problems/simplify-path

Java Solution: github

package leetcode;/* * 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". */public class P071_SimplifyPath {/* * 建立最简单的树形结构 * 6 ms */static class Solution1 {    public String simplifyPath(String path) {    int len = 0;    if (path == null || (len = path.length()) < 2)    return path;    Node head = new Node(), cur = head;    StringBuilder st = new StringBuilder();    boolean isLastSlash = true;    boolean isLastDot = false;    boolean isDoubleDot = false;    boolean isDoubleDotLast = false;    for (int i = 0; i <= len; i ++) {    char c = '\0';    if (i == len || (c = path.charAt(i)) == '/') {    if (isLastSlash || (isLastDot && ! isDoubleDot)) {    if (st.length() != 0)    st = new StringBuilder();    isLastSlash = true;    isDoubleDot = false;    continue;    }    if (isDoubleDot) {    if (st.length() == 2) {    if (i != len) {    cur = cur.pre == null ? cur : cur.pre;    isDoubleDotLast = true;    } else {    if (cur.pre != null)    cur.pre.next = null;    }    } else {    if (! st.toString().equals(".")) {    Node temp = new Node();        temp.str = st.toString();        temp.pre = cur;        cur.next = temp;        cur = temp;    }    }    st = new StringBuilder();    isDoubleDot = false;    continue;    }    if (! st.toString().equals(".")) {    Node temp = new Node();    temp.str = st.toString();    temp.pre = cur;    cur.next = temp;    cur = temp;    }    st = new StringBuilder();    isDoubleDot = false;    isLastDot = false;    isLastSlash = true;    } else if (c == '.') {    if (isLastDot)    isDoubleDot = true;    st.append(".");    isLastDot = true;    isLastSlash = false;    isDoubleDotLast = false;    } else {    st.append(c);    isDoubleDot = false;    isLastDot = false;    isLastSlash = false;    isDoubleDotLast = false;    }    }    if (isDoubleDotLast && cur != null)    cur.next = null;    if (cur.str == null)    cur.next = null;    if (cur.next != null)    cur.next = null;    Node travel = head;    StringBuilder ans = new StringBuilder();    while (travel != null) {    ans.append(travel.str == null ? "/" : travel.str);    if (travel != head && travel.next != null)    ans.append("/");    travel = travel.next;    }return ans.toString();    }    static class Node {    String str = null;    Node pre;    Node next;    public Node() {}    public Node(String str, Node pre, Node next) {    this.str = str;    this.pre = pre;    this.next = next;    }    }}}

C Solution: github

/*    url: leetcode.com/problems/simplify-path    AC 6ms 28.00%*/#include <stdio.h>#include <stdlib.h>//element typetypedef int T;typedef struct dll sdll;typedef struct dll * pdll;typedef struct dln sdln;typedef struct dln * pdln;//doubly list nodestruct dln {    T val;    pdln pre;    pdln nxt;};//doubly linked liststruct dll {    pdln first;    pdln last;    int size;};pdll dll_init() {    pdll l = (pdll) malloc(sizeof(sdll));    l->first = NULL;    l->last = NULL;    l->size = 0;    return l;}void dll_add_first(pdll l, T v) {    pdln t = (pdln) malloc(sizeof(sdln));    t->val = v;    t->pre = NULL;    t->nxt = NULL;    if (l->size == 0) {        l->first = t;        l->last = t;        l->size = 1;        return;    }    t->nxt = l->first;    l->first->pre = t;    l->first = t;    l->size ++;}void dll_add_last(pdll l, T v) {    pdln t = (pdln) malloc(sizeof(sdln));    t->val = v;    t->pre = NULL;    t->nxt = NULL;    if (l->size == 0) {        l->first = t;        l->last = t;        l->size = 1;        return;    }    t->pre = l->last;    l->last->nxt = t;    l->last = t;    l->size ++;}void dll_remove_first(pdll l) {    pdln t = NULL;    if (l->first == NULL) return;    if (l->first == l->last) {        free(l->first);        l->first = NULL;        l->last = NULL;        l->size = 0;        return;    }    t = l->first->nxt;    t->pre = NULL;    free(l->first);    l->first = t;    l->size --;}void dll_remove_last(pdll l) {    pdln t = NULL;    if (l->last == NULL) return;    if (l->first == l->last) {        free(l->first);        l->first = NULL;        l->last = NULL;        l->size = 0;        return;    }    t = l->last->pre;    t->nxt = NULL;    free(l->last);    l->last = t;    l->size --;}T* dll_convert_to_array(pdll l) {    T* arr = NULL;    int arr_index = 0;    pdln travel = NULL;    pdln t1 = NULL, t2 = NULL;    if (l == NULL || l->size == 0) return arr;    arr = (T*) malloc(sizeof(T) * l->size);    travel = l->first;    while (travel != NULL) {        arr[arr_index ++] = travel->val;        travel = travel->nxt;    }    t1 = l->first;    while (t1 != NULL) {        t2 = t1->nxt;        free(t1);        t1 = t2;    }    free(l);    return arr;}void dll_print(pdll l) {    pdln t = l == NULL ? NULL : l->first;    while (t != NULL) {        printf("%d ", t->val);        t = t->nxt;    }    printf("\r\n");}void dll_free_all(pdll l) {    pdln t1 = l->first, t2 = NULL;    while (t1 != NULL) {        t2 = t1->nxt;        free(t1);        t1 = t2;    }    free(l);}char* simplifyPath(char* p) {    pdll l = dll_init();    int pn = 0, i = 0, sti = 0, len = 0, ai = 0;    char c = '\0', *ans = NULL;     pdln n = NULL;    while (1) {        c = p[i ++];        if (c == '\0') break;        if (c == '/') {            if (len == 2 && p[sti] == '.' && p[sti+1] == '.') {                dll_remove_last(l);            } else if (len != 0 && ! (len == 1 && p[sti] == '.')) {                dll_add_last(l, sti);            }            sti = i;            len = 0;            continue;        }        len ++;    }    if (len == 2 && p[sti] == '.' && p[sti+1] == '.') {        dll_remove_last(l);    } else if (len != 0 && ! (len == 1 && p[sti] == '.')) {        dll_add_last(l, sti);    }    pn = i;    ans = (char*) malloc(sizeof(char) * pn);    n = l->first;    while (n != NULL) {        i = n->val;        ans[ai ++] = '/';        while (1) {            c = p[i++];            if (c == '\0' || c == '/') break;            ans[ai ++] = c;        }        n = n->nxt;    }    ans[ai ++] = '\0';    if (l->first == NULL) {        ans[0] = '/';        ans[1] = '\0';        return ans;    }    return ans;}int main() {    char* p = "/home//foo/";    char* a = simplifyPath(p);    printf("answer is %s\r\n", a);    free(a);    return 0;}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/simplify-path/    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年4月16日    @details:    Solution: 65ms 36.88%'''class Solution(object):    def simplifyPath(self, p):        """        :type p: str        :rtype: str        """        ps = p.split("/")        l = []        for s in ps:            if s == "": continue            if s == ".": continue            if s == "..":                 if len(l) > 0: l.pop()                continue            l.append(s)        if len(l) == 0:            return "/"        ans = ""        for s in l:            ans += "/"            ans += s        return ansif __name__ == "__main__":    p = "/"    print(Solution().simplifyPath(p))

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