Create simplify_path.md

Author: 17020021029

simplify_path.md

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+# Simplify Path
+## Problem 
+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".
+## 问题分析
+主要是对数组遍历分情况对数组inking处理，
+## 代码实现
+```C
+char* simplifyPath(char* path) {
+    int i, t=1;//i用做数组下标，t用来计算返回路径的下标
+	int len = strlen(path);
+	path[0] = '/';  //返回的路径第一个都为'/'
+    for(i=1; i<len; i++) {
+        if(path[i]=='.' && ((i+1<len && path[i+1]=='/') || path[i+1]=='\0')) {
+            i = i + 1;
+        } 
+        else if(path[i]=='.' && i+1<len && path[i+1]=='.' && ((i+2<len && path[i+2]=='/') || path[i+2]=='\0')) {
+            i = i + 2;
+            t = t - 2 < 0? 0: t-2;
+            while(path[t]!='/') {
+                t--;
+            }
+			t++;
+        } 
+        else if(path[i]=='/') {
+			continue;
+		} 
+    else {
+            while(i<len && path[i]!='/'){
+                path[t++] = path[i++];
+            } 
+            path[t++] = path[i];
+        }
+    }
+	if(t>1) {
+		path[t-1] = '\0';
+	} 
+    else {
+		path[t] = '\0';
+	}
+	return path;
+}
+
+```