Determines whether a group of rectangles compose a complete rectangle.

Determines whether a group of rectangles compose a complete rectangle.

function rect(l, t, w, h) {    this.l = l; // left    this.t = t; // top    this.w = w; // width    this.h = h; // height}function check(rects) {    var data = [];    var i, t, b;    for(i = 0; i < rects.length; i++ ) {        t = rects[i].t;        b = t + rects[i] + h;        while (t < b) {            if (!data[t]) {                data[t] = {                    l: rects[i].l,                    w: rects[i].w                }            } else {                data[t].l = Math.min(data[t].l, rects[i].l);                data[t].w += rects[i].w;            }            t++;        }    }    for (i = 1; i < data.length; i++) {        if (!data[i] || (data[i].l != data[i - 1].l) || (data[i].w != data[i - 1].w)) return false;    }    return true;}

The time complexity of this method is O (h1 + h2 +... + Hn), which can be generally understood as O (n * avg (h). If there are many rectangles, but each height is relatively small, it is similar to O (n ), however, if there are few Rectangles and the height is very high, the efficiency is worse than O (n ^ 2.

There is another method below

function check(rects) {    var i, l = 10000, r = 0, t = 10000, b = 0, sum = 0;    for (i = 0; i < rects.length; i++) {        sum += rects[i].w * rects[i].h;        l = Math.min(l, rects[i].l);        r = Math.max(r, rects[i].l + rects[i].w);        t = Math.min(t, rects[i].t);        b = Math.max(b, rects[i].t + rects[i].h);    }    return ((r - l) * (b - t) != sum);}

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