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License: MIT License
无线前端库
License: MIT License
function quadratic2cubicBezier(a, b) {
return [[(a / 3 + (a + b) / 3 - a) / (b - a), (a * a / 3 + a * b * 2 / 3 - a * a) / (b * b - a * a)],
[(b / 3 + (a + b) / 3 - a) / (b - a), (b * b / 3 + a * b * 2 / 3 - a * a) / (b * b - a * a)]];
}
return quadratic2cubicBezier(this.v / this.a , this.t + this.v / this.a);
ml.js 中 function getCommonAncestor(el1, el2)函数
两个节点先求深度,深度大的向上走到深度小的节点同一深度。。。然后两个节点一起往上走,直到相遇。这样算法复杂度是否能小一点?
每次contains判断,(貌似)需要遍历整个子树,增加不必要的开销。
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