The title
!!!!! Simplified path – force link
Analysis of the
This may seem complicated at first glance, but a closer look at a few examples reveals that there aren’t really many cases to deal with, except:
null..path
What to do with them? Compared to the valid parentheses problem, the first thing you can think of is using stacks to help us solve the problem:
- encounter
null或., do not do processing - encounter
pathThat will bepathInto the stack - encounter
.And out of the stack
This way, the code is easy to implement:
var simplifyPath = function (path) {
const stack = [];
const pathArr = path.split("/");
for (let i = 0; i < pathArr.length; i++) {
const p = pathArr[i];
if (p === "..") {
stack.pop();
} else if (p === "." || p === "") {
continue;
} else {
stack.push(p);
}
}
return "/" + stack.join("/");
};
Copy the codeThe results are as follows:
To optimize the
The above code does not look too ugly because there is not much logic to deal with. If we need to deal with additional logic, for example, if we encounter ~ and think it is /home/user, we will have to add additional conditional branches, and the code will gradually become big and messy and difficult to maintain.
Here we can use the strategy pattern to optimize:
var simplifyPath = function (path) {
const stack = [];
const pathArr = path.split("/");
const strategy = {
"": stack= > stack,
".": stack= > stack,
"..": stack= > stack.pop()
};
for (let i = 0; i < pathArr.length; i++) {
const p = pathArr[i];
if (p in strategy) {
strategy[p](stack);
} else {
stack.push(p);
}
}
return "/" + stack.join("/");
};
Copy the codeThe results are as follows:
