我已经实现了一种算法,用于将重叠的现有范围划分为日期/数字范围列表。
算法正在运行,但我想知道你是否可以避免算法的最后一个循环。
有可能吗?
0-100(红色) 90-150(绿色) 90-150(蓝色) 140-300(黄色) 170-240(黑色) 350-530(橙色) 50-500(银) 50-60(粉红色)
0-49(红色) 50-60(红色,银色,粉色) 61-89(红色,银色) 90-100(红,绿,蓝,银) 101-139(绿,蓝,银) 140-150(绿,蓝,黄,银) 151-169(黄色,银) 170-240(黄色,黑色,银色) 241-300(黄色,银) 301-349(银) 350-500(橙,银) 501-530(橙色)
Javascript代码:
function splitRanges(original_intervals) {
for (var to = [], from = [], n, i = original_intervals.length; i--;) {
if (to.indexOf(n = original_intervals[i].to) < 0)
to.push(n);
if (from.indexOf(n = original_intervals[i].from) < 0)
from.push(n);
}
to.sort(function(a, b) {
return a - b;
});
from.sort(function(a, b) {
return a - b;
});
var intervals = [];
while (to.length) {
var sFrom = from.shift();
var sTo = 0;
if (from.length == 0) {
sTo = (from.push((n = to.shift()) + 1), n);
} else {
if (from[0] > to[0]) {
while (to[0] < from[0]) {
from.unshift(to[0] + 1);
to.shift();
}
sTo = from[0] - 1;
} else {
sTo = from[0] - 1;
}
}
intervals.push({
from: sFrom,
to: sTo,
colors: []
});
}
/***********************Loop that i want remove*/
intervals.forEach(function(item, index) {
original_intervals.forEach(function(item1, index1) {
if ((item.from >= item1.from && item.from <= item1.to) || (item.to >= item1.from && item.to <= item1.to))
item.colors.push(item1.color);
});
});
return intervals;
}
var r1 = [{
id: 1,
from: 0,
to: 100,
color:'red'
}, {
id: 2,
from: 90,
to: 150,
color:'green'
}, {
id: 3,
from: 90,
to: 150,
color:'blue'
}, {
id: 4,
from: 140,
to: 300,
color:'yellow'
}, {
id: 5,
from: 170,
to: 240,
color:'black'
}, {
id: 6,
from: 350,
to: 530,
color:'orange'
}, {
id: 7,
from: 50,
to: 500,
color:'silver'
}
, {
id: 8,
from: 50,
to: 60,
color:'pink'
}
];
console.log(splitRanges(r1));
答案 0 :(得分:0)
你需要一些迭代,至少是获得所有范围点的迭代,然后从中生成一个数组并为每个小间隔获取颜色子集。
public ResponseEntity<?> create(
@RequestBody AttributeDTO policy,
BindingResult bindingResult)
{
logger.debug("CreatePolicy=" + policy);
...
var data = [{ from: 0, to: 100, color: 'red' }, { from: 90, to: 150, color: 'green' }, { from: 90, to: 150, color: 'blue' }, { from: 140, to: 300, color: 'yellow' }, { from: 170, to: 240, color: 'black' }, { from: 350, to: 530, color: 'orange' }, { from: 50, to: 500, color: 'silver' }, { from: 50, to: 60, color: 'pink' }],
ranges = new Set,
parts,
result;
data.forEach(({ from, to }) => (ranges.add(from), ranges.add(to)));
parts = [...ranges].sort((a, b) => a - b);
result = parts.slice(1).map(function (a, i, aa) {
var from = i ? aa[i - 1] : parts[0],
to = a,
colors = data.filter(d => d.from <= from && to <= d.to).map(({ color }) => color);
return { from, to, colors };
});
console.log(result);