使用javascript在数组中合并具有间隔属性的重叠对象

Question

我发现了类似的问题，但批准的答案不适用于我的问题。

我有一个输入：范围对象数组，每个对象包含：

• start：整数，范围的开始，
• end：整数，范围结束。

输出应该是：

覆盖与输入相同范围的非重叠范围对象数组，按从最小开始到最大开始的顺序排列。如果满足以下条件，则两个范围不重叠：

• range1.start <= range2.start 和
• range1.end >= range2.start

输入：

[
  { start: 8, end: 10 },
  { start: 5, end: 7  },
  { start: 9, end: 12 },
  { start: 2, end: 6  },
]

输出：

 [
   { start: 2, end: 7  },
   { start: 8, end: 12 }
 ]

正如我所提到的，我尝试在网络上应用解决方案来合并重叠间隔，但它们没有完成这项工作。

谢谢。

Answer 1

您可以按start 和end 对数组进行排序，并通过检查范围是否重叠来迭代排序后的数组。

var data = [{ start: 8, end: 10 }, { start: 5, end: 7 }, { start: 9, end: 12 }, { start: 2, end: 6 }],
    result = data
        .sort(function (a, b) { return a.start - b.start || a.end - b.end; })
        .reduce(function (r, a) {
            var last = r[r.length - 1] || [];
            if (last.start <= a.start && a.start <= last.end) {
                if (last.end < a.end) {
                    last.end = a.end;
                }
                return r;
            }
            return r.concat(a);
        }, []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

var ranges = [
  { start: 8, end: 10 },
  { start: 5, end: 7  },
  { start: 9, end: 12 },
  { start: 2, end: 6  }
];


function merge(ranges) {
  // first, sort the ranges
  ranges.sort((a, b) => a.start - b.start);
  
  // take two ranges, and merges them together
  var mergeFn = (a, b) => ({start: Math.min(a.start, b.start), end: Math.max(a.end, b.end)});
  
  // check if two ranges overlap
  var overlapFn = (a, b) => (b.start <= a.end);
  
  // make current the first item of the array (start the array from 1 to not check the first item against itself)
  var current = ranges[0];
  var result = [];
  for(var i = 1; i < ranges.length; i++) {
    if(overlapFn(current, ranges[i])) // if the current range overlapping with this range
       current = mergeFn(current, ranges[i]); // merge them into the current range
    else { // if not
      result.push(current); // add the current accumulated range as result
      current = ranges[i]; // start accumulating another one from this range
    }
  }
  result.push(current); // add the last result

  return result;
}

console.log(merge(ranges));