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[英]jQuery Border Image (top-left, top-right, bottom-left, bottom-right)
[英]jquery: sort 4 points to (top-left, top-right, bottom-right, bottom-left)
給定:2個數組,每個數組包含4個點的坐標:
point_array_a = [point_a_1_x, point_a_1_y, point_a_2_x, ..., point_a_4_y]
point_array_b = [point_b_1_x, ... ..., point_b_4_y]
任務:
對point_array_a進行排序,以便最后按以下順序列出點:
point_array_a_sorted = [top-left_x, top_left_y, top-right_x, top-right_y, bottom-right_x, bottom_right_y, bottom-left_x, bottom_left_y]
以相同的方式對point_array_b進行排序,以使point_a_k_l對應於開始時的point_b_k_l。
恐怕沒有簡單的算法可以做到這一點。 但是下面的代碼片段就可以了(假設y坐標較大的點低於y坐標較低的點):
var i, points = [], leftX = point_array_a[0], topY = point_array_a[1];
for (i = 0; i < 4; i++)
{
leftX = Math.min(leftX, point_array_a[i * 2]);
topY = Math.min(topY, point_array_b[i * 2]);
points.push([
[point_array_a[i * 2], point_array_a[i * 2 + 1]],
[point_array_b[i * 2], point_array_b[i * 2 + 1]]
]);
}
points.sort(function(first, second){
if (first[0][0] == leftX)
return first[0][1] == topY ? -1 : 1;
if (second[0][0] == leftX)
return second[0][1] == topY ? 1 : -1;
return first[0][1] < second[0][1] ? -1 : 1;
});
var point_array_a_sorted = [], point_array_b_sorted = [];
for (i = 0; i < 4; i++)
{
point_array_a_sorted.push(points[i][0][0], points[i][0][1]);
point_array_b_sorted.push(points[i][1][0], points[i][1][1]);
}
我們通過為現有的Array.sort
函數提供正確的對象來進行比較和交換(成對的點)來利用它們。
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