我很难解决游戏板问题。我能够迭代我的辅助函数一次，然后它就会停止。我很困惑为什么我的 while 循环不会获取当前的答案，添加到当前的解决方案并在板上画一个 X。我尝试过很多不同的答案。我需要 for 循环多次循环，并且板子需要在原来的位置用 X 进行更新。我还需要队列和访问过的数组来更新
这应该是通过 findpath 函数到达末端节点的最短第一路径函数。我在棋盘上的骑士棋子的所有可能方向上循环，直到到达最终位置或者队列中的空间或物品用完
let board = [[], [], [], [], [], [], [], []];
queue = [];
visited = [];
let counter = 0;
let xPossibleMoves = [2, 1, -1, -2, -2, -1, 1, 2];
let yPossibleMoves = [1, 2, 2, 1, -1, -2, -2, -1];
function findPath(currentX, currentY, endX, endY) {
console.log(currentX, currentY);
console.log(queue);
console.log(visited);
queue.forEach(function (item) {
visited.push(item);
});
console.log(visited);
for (let i = 0; i < 8; i++) {
let vectorX = currentX + xPossibleMoves[i];
let vectorY = currentY + yPossibleMoves[i];
console.log("next hop X: " + vectorX + " Y: " + vectorY);
if (board[vectorX] == undefined || board[vectorY] == undefined) continue;
if (board[vectorX][vectorY] == "X") continue;
if (board[vectorX][vectorY] == "END") return "DONE";
queue.push(vectorX, vectorY);
board[vectorX][vectorY] = "X";
}
//Stuff out of my for loop happens Once
tail = 2;
queue = queue.slice(tail);
console.log(queue + " : queue");
console.log(board);
console.log(visited);
queue.forEach(function (item) {
visited.push(item);
});
console.log(visited);
//causes a infinite loop
//while (board[[endX][endY]] == null)
//findPath(currentX, currentY, endX, endY);
}
function knightsMove(startX, startY, endX, endY) {
if (board[endX][endY] == null) {
let currentX = startX;
let currentY = startY;
queue.push(currentX);
queue.push(currentY);
board[endX][endY] = "END";
board[currentX][currentY] = "START";
console.log(board);
findPath(currentX, currentY, endX, endY);
}
}
knightsMove(0, 0, 3, 3);
我尝试添加 do while 循环、while 循环、制作原始实例的原型、大量不同的变体、修改任何会阻止循环的错误、添加一个函数来更新所有内容并推送 while 循环等等。
嗯...我不想做你的全部作业，但我在 Chrome 中保留了一些旧的片段。这至少应该有助于清理你的输出并使调试更容易。我也知道这种感觉——你最终得到的
console.log(...)console.clear();
// This is just a variable to help differentiate the console output lines.
let dumpInfoCount = 0;
// Some console styles to make reading the output a little easier...
cs = {
h1: "font-size:150%;",
h2: "font-size:130%;",
i: "font-style: italic;",
b: "font-weight:bold;",
attn: "color: #990000;",
warn: "color: #ffcc00;",
good: "color: #009900;",
info: "color: #00ccff;",
gray: "color: #ccc;",
reset: "color: inherit;font-size:100%;",
};
let board = [[], [], [], [], [], [], [], []];
queue = [];
visited = [];
let counter = 0;
let xPossibleMoves = [2, 1, -1, -2, -2, -1, 1, 2];
let yPossibleMoves = [1, 2, 2, 1, -1, -2, -2, -1];
/**
* Clean up code by removing all the `console.log(...)`
* lines. Also makes it easier to turn it all off later.
*/
function dumpInfo(cx, cy, extraMessage) {
console.groupCollapsed(
"%cInfo Number:%c %d%c %s",
cs.warn,
cs.reset,
dumpInfoCount,
cs.info,
extraMessage ? extraMessage : ""
);
console.log(
"%cPassed In %cX%c,%cY:%c %d , %d",
cs.info,
cs.good,
cs.reset,
cs.good,
cs.reset,
cx,
cy
);
console.log("%cBoard:%c %o", cs.info, cs.reset, board);
console.log("%cQueue:%c %o", cs.info, cs.reset, queue);
console.log("%cVisited:%c %o", cs.info, cs.reset, visited);
console.groupEnd();
dumpInfoCount += 1;
}
function findPath(currentX, currentY, endX, endY) {
dumpInfo(currentX, currentY);
// Not sure why the need to clone into `visited` here. Nothing happens with it.
// queue.forEach(function (item) {
// visited.push(item);
// });
// dumpInfo(currentX, currentY);
console.groupCollapsed("%cInner Loop Here", cs.info);
for (let i = 0; i < 8; i++) {
let vectorX = currentX + xPossibleMoves[i];
let vectorY = currentY + yPossibleMoves[i];
// Just be mindful that we're passing a differnt X/Y pair, here...
dumpInfo(vectorX, vectorY, "Inner Loop Here");
if (board[vectorX] == undefined || board[vectorY] == undefined) continue;
if (board[vectorX][vectorY] == "X") continue;
if (board[vectorX][vectorY] == "END") return "DONE";
queue.push(vectorX, vectorY);
board[vectorX][vectorY] = "X";
}
console.groupEnd();
//Stuff out of my for loop happens Once
tail = 2;
queue = queue.slice(tail);
dumpInfo(currentX, currentY, "BEFORE Queue to Visited");
queue.forEach(function (item) {
visited.push(item);
});
dumpInfo(currentX, currentY, "AFTER Queue to Visited");
queue.forEach((item) => {
findPath(currentX, currentY, endX, endY);
});
// while (queue.length > 0 && board[[endX][endY]] == null)
// return findPath(currentX, currentY, endX, endY);
/**
* Need to decide what the final return value is. It's not
* clear how you're defining "success", here.
*/
return { currentX, currentY, endX, endY };
}
function knightsMove(startX, startY, endX, endY) {
if (board[endX][endY] == null) {
// let currentX = startX;
// let currentY = startY;
queue.push(startX);
queue.push(startY);
board[endX][endY] = "END";
board[startX][startY] = "START";
dumpInfo(startX, startY);
return findPath(startX, startY, endX, endY);
}
}
knightsMove(0, 0, 3, 3);
我曾经使用过一个很酷的小型 Chrome 扩展程序，名为 jsShell，但那家伙不再支持它了。现在我使用一种叫做 Snippets 的东西。我知道它没有任何自动化功能（比如
run this script if the URL matches this regex祝你好运！