A long time ago, there was a road leading to the unreachable west...
A young cowboy, according to the instructions, had to go from one place to another. Instructions like this"NORTH", "SOUTH", "WEST", "EAST".
Obviously, "NORTH" and "SOUTH" are in opposite directions, "WEST" and "EAST" is also the opposite.
Walking in one direction and then going back is undoubtedly a useless effort.
In this inaccessible western wilderness, with bad weather and scarce water resources, it is important to save some energy, otherwise you may die!
How to take the wisest route is very important!
Instructions for cowboys, like this:
plan = ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]
You can see at a glance, "NORTH" and "SOUTH", this route is obviously unreasonable, it is best to stay where you are.
So, your task is how to streamline the route to save energy.
A better solution would be like this:
plan = ["WEST"]
Another example:
["NORTH", " SOUTH", "EAST", "WEST"]
Here, "NORTH" and "SOUTH" are offset, "EAST " and "WEST" are offset, and finally an empty array is returned.
Let’s look at a more complicated example:
["NORTH", "EAST", "WEST", "SOUTH", "WEST", "WEST"]
"EAST", "WEST" is offset to get ["NORTH", "SOUTH", "WEST", "WEST"]
"NORTH", "SOUTH" are offset, and finally we get ["WEST", "WEST"].
However, please note that the following situations cannot be offset:
["NORTH", "WEST", "SOUTH", "EAST"]
Because "EAST", "WEST" or "NORTH", "SOUTH" are not adjacent, but separated.
Let’s take a look at how to write such a route reduction function.
It accepts a string array as a parameter and returns a new string array.
Like this:
dirReduc(["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"]) // ["WEST"] dirReduc(["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH"]) // []
The first step is to establish the mapping relationship of the directions, which are the opposite directions:
var opposite = { "NORTH":"SOUTH", "SOUTH":"NORTH", "EAST":"WEST", "WEST":"EAST" };
Next, traverse the string array plan from back to front, and when encountering the opposite direction, remove it from the array and offset it.
Then, start the second round of traversal, removal, offset, third round, Nth round, until you can't find the opposite direction and jump out of the loop.
The array at this time is the streamlined best route.
function dirReduc(arr){ var flag = false; while(!flag){ for(var i=arr.length-2,flag=true;i>=0;i--){ if(opposite[arr[i]] === arr[i+1]){ arr.splice(i+1,1); arr.splice(i,1); i--; flag = false; } } } return arr; }
The above is the content of JavaScript interesting question: The Lost Cowboy. For more related content, please pay attention to the PHP Chinese website (www.php.cn)!