效果

思路
- canvans 绘制棋盘,绘制时候边缘预留棋子位置
- 监听点击事件绘制落子并记录到字典中
- 获胜判定,在四个方向上检测是否有足够数量的连贯棋子

代码
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>ym</title>
<style>
canvas {
display: block;
margin: 0 auto;
border: 0
}
.result {
text-align: center;
}
button{
display: block;
margin: 0 auto;
padding: 4px 20px;
border:0;
color: #fff;
outline: none;
border-radius: 3px;
background: #43a6ff
}
button:hover{
font-weight: bold;
cursor: pointer;
}
</style>
</head>
<body>
<canvas id="cv" width="200px" height="200px"></canvas>
<p class="result"></p>
<button onclick="loadPanel(400, 400,30,13)">刷新</button>
<script>
loadPanel(400, 400,30,13);
function loadPanel(w, h, cs, ps) {
let i, j, k;
let chks = [[1, 0], [0, 1], [1, 1], [1, -1]];
let successNum = 5;
let resultEl = document.querySelector('.result');
cs = cs || 16;
ps = ps || 4;
h = h || w;
let el = document.getElementById('cv');
el.setAttribute('width', w + 'px');
el.setAttribute('height', h + 'px');
let context = el.getContext("2d");
let splitX = ~~((w - 2 * ps) / cs), splitY = ~~((h - 2 * ps) / cs);
let pieces = {};
context.translate(ps, ps);
context.beginPath();
context.strokeStyle = '#000';
for (i = 0; i < splitX + 1; i++) {
context.moveTo(cs * i, 0);
context.lineTo(cs * i, splitY * cs);
context.stroke();
}
for (j = 0; j < splitY + 1; j++) {
context.moveTo(0, cs * j);
context.lineTo(splitX * cs, cs * j);
context.stroke();
}
context.closePath();
let user = 0, colors = ['#000', '#fefefe'];
el.addEventListener('click', function (e) {
let x = e.offsetX,
y = e.offsetY,
rx = ~~((x - ps) / cs) + (((x - ps) % cs <= cs / 2) ? 0 : 1),
ry = ~~((y - ps) / cs) + (((y - ps) % cs <= cs / 2) ? 0 : 1);
context.beginPath();
context.arc(cs * rx, cs * ry, ps, 2 * Math.PI, false);
context.fillStyle = colors[user];
context.strokeStyle = '#000';
user ? user = 0 : user = 1;
context.fill();
context.stroke();
context.closePath();
let piece = pieces[rx + '-' + ry] = user;
for (j = 0; j < chks.length; j++) {
let num = 1, chk = chks[j];
for (i = 1; i <= 4; i++) {
if (pieces[(rx + chk[0] * i) + '-' + (ry + chk[1] * i)] == piece) {
num++
} else {
for (i = -1; i >= -4; i--) {
if (pieces[(rx + chk[0] * i) + '-' + (ry + chk[1] * i)] == piece) {
num++
}
}
break
}
}
if (num == successNum) {
resultEl.innerHTML = ['白', '黑'][user] + '方赢';
break;
}
}
})
}
</script>
</body>
</html>