方向缩减挑战要求您从方向数组中找到最短路线。
Input -> Output ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"] -> ["WEST"] /* Because moving "NORTH" and "SOUTH", or "EAST" and "WEST", is unnecessary. */ /* Case: omit the first "NORTH"-"SOUTH" pair ["NORTH", "SOUTH", "SOUTH", "EAST", "WEST", "NORTH", "WEST"] -> ["SOUTH", "EAST", "WEST", "NORTH", "WEST"] Case: omit the "EAST"-"WEST" pair -> ["SOUTH", "NORTH", "WEST"] Case: omit the "NORTH"-"SOUTH" pair -> ["WEST"] */ // this case cannot be reduced: ["NORTH", "WEST", "SOUTH", "EAST"] -> ["NORTH", "WEST", "SOUTH", "EAST"]
function dirReduc(plan) {
var opposite = {
'NORTH': 'SOUTH', 'EAST': 'WEST', 'SOUTH': 'NORTH', 'WEST': 'EAST'};
return plan.reduce(function(dirs, dir){
if (dirs[dirs.length - 1] === opposite[dir])
dirs.pop(); // Remove last direction if it's the opposite
else
dirs.push(dir); // Otherwise, add current direction to the stack
return dirs;
}, []);
}
这是使用reduce实现的LIFO（后进先出）方法。
function dirReduce(arr) {
let str = arr.join(''), pattern = /NORTHSOUTH|EASTWEST|SOUTHNORTH|WESTEAST/;
while (pattern.test(str)) str = str.replace(pattern,''); // Remove pairs while they exist
return str.match(/(NORTH|SOUTH|EAST|WEST)/g)||[];
}
替代版本：
function dirReduc(arr) {
const pattern = /NORTHSOUTH|EASTWEST|SOUTHNORTH|WESTEAST/;
let str = arr.join('');
while (pattern.test(str)) {
str = str.replace(pattern, ''); // Remove pairs while they exist
}
const result = str.match(/(NORTH|SOUTH|EAST|WEST)/g);
return result || [];
}
我认为该解决方案的最低情况如下：
1. ["SOUTH", "EAST", "WEST", "NORTH"] -> [] 2. ["NORTH", "WEST", "SOUTH", "EAST"] -> ["NORTH", "WEST", "SOUTH", "EAST"]
我更喜欢解决方案2，因为它简洁并且巧妙地使用了正则表达式。最初，我无法想象用正则表达式来解决它，但是使用 join、match、while、replace 和 test 方法来消除对是令人印象深刻的。
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