javascript - 在javascript中按关系对数组进行排序

Question

我想对一个数组进行排序。

数组中的项是有关系的。

例如。 list[5] 应该在 list[9] 之前 但 在 list[3] 之后

---

样本中的期望值仅供测试。它实际上并不存在。

---

这是一个包含关系和预期索引的示例数组。

var list = [{
  id: '0001',
  before: '0002',
  expected: 0
}, {
  id: '0002',
  before: '0007',
  after: '0001',
  expected: 4
}, {
  id: '0003',
  before: '0006',
  after: '0001',
  expected: 2
}, {
  id: '0004',
  after: '0007',
  expected: 11
}, {
  id: '0005',
  before: '0003',
  after: '0001',
  expected: 1
}, {
  id: '0006',
  before: '0002',
  after: '0001',
  expected: 3
}, {
  id: '0007',
  before: '00010',
  after: '0002',
  expected: 5
}, {
  id: '0008',
  before: '00012',
  after: '0007',
  expected: 9
}, {
  id: '0009',
  before: '0011',
  after: '0001',
  expected: 7
}, {
  id: '0010',
  before: '0009',
  after: '0007',
  expected: 6
}, {
  id: '0011',
  before: '0008',
  after: '0001',
  expected: 8
}, {
  id: '0012',
  before: '0004',
  after: '0010',
  expected: 10
}];

Answer 1

关于 Jan Turoň 的扩展:这里是一个示例，说明您的问题(在有向无环图中找到全序)如何无法解决:

var list = [{ id:A, before:B }, // "First" in total order
            { id:B, after:A }, // "Last" in total order
            { id:C, after:A, before:B },
            { id:D, after:A, before:B }];

C 和 D 之间没有总序:你可以称它们相等，但如果你有一个列表而不是 D 会怎样D0 -> D1 -> D2?

根据您的问题类型，这可以通过预处理来解决:将 2 级节点的路径减少为单个节点，并调用通过 2 级相同节点的并行路径(也将减少为单个节点)。在这样的预处理结束时，您会留下一棵树 - 在您的情况下应该是一个列表/路径(或单个节点，因为您减少了 2 度顶点的路径)。

请注意，“之前”和“之后”的信息是多余的:您实际上只需要其中之一。语句“A 在 B 之前”等同于“B 在 A 之后”，您的无环图只需要反射(reflect)“之前”或“之后”的方向。然后您正在寻找的是包含所有节点的图形路径(因为它们被定向到“之前”或“之后”，您会自动按顺序从头到尾获得路径 - 如果存在这样的路径):

// First build the adjacency list for "X before Y"
var befores = { };
for(var i = 0; i < count; ++i)
    befores[list[i].id] = null;
function insert(before, after) {
    if(!before || !after)
        return;
    befores[before] = { next:befores[before], id:after };
}
for(var i = 0; i < count; ++i) {
    var item = list[i];
    insert(item.after, item.id); // "X after Y" -> "Y before X"
    insert(item.id, item.before);
}

// build complete the graph as a lookup table
// id is "before" all elements in lookup[id]
var lookup = { };
for(var i = 0; i < count; ++i) {
    var id = list[i].id;
    var beforeList = [id];
    var beforeSet = { };
    beforeSet[id] = 1;
    // use "A before B" and "B before C" to gain "A before C"
    for(var j = 0; j < beforeList.length; ++j) {
        for(var item = befores[beforeList[j]]; item != null; item = item.next) {
            if(!beforeSet[item.id]) {
                beforeList.push(item.id);
                beforeSet[item.id] = 1;
            }
        }
    }
    // for our comparison we don't care if id is in beforeSet
    lookup[id] = beforeSet;
    // slice beforeList to get only the elements after id here:
    //beforeList = beforeList.slice(1, beforeList.length);
}

// Now sort using the following
// a) if rhs is present in "before"-set of lhs then lhs < rhs
// b) if rhs is not present then rhs < lhs
// c) there is information missing from our graph if a) and b) for lhs analogous lead to a different conclusion!
list.sort(function(lhs, rhs) {
    if(!lhs.after || !rhs.before) return -1;
    if(!lhs.before || !rhs.after) return 1;
    if(lhs.id === rhs.id) return 0;
    // different ids guaranteed, doesn't matter if lookup[id] contains id itself
    var result = lookup[lhs.id][rhs.id] ? -1 : 1;
    // expect reversing lhs and rhs to get inverse result
    var expected = lookup[rhs.id][lhs.id] ? 1 : -1;
    if(result != expected) {
        // alert: there is no adjacency information between lhs and rhs!
    }
    return result;
});

自己测试一下:

var list = [{
  id: '0001',  before: '0002',  expected: 0}, {
  id: '0002',  before: '0007',  after: '0001',  expected: 4}, {
  id: '0003',  before: '0006',  after: '0001',  expected: 2}, {
  id: '0004',  after: '0007',  expected: 11}, {
  id: '0005',  before: '0003',  after: '0001',  expected: 1}, {
  id: '0006',  before: '0002',  after: '0001',  expected: 3}, {
  id: '0007',  before: '0010',  after: '0002',  expected: 5}, {
  id: '0008',  before: '0012',  after: '0007',  expected: 9}, {
  id: '0009',  before: '0011',  after: '0001',  expected: 7}, {
  id: '0010',  before: '0009',  after: '0007',  expected: 6}, {
  id: '0011',  before: '0008',  after: '0001',  expected: 8}, {
  id: '0012',  before: '0004',  after: '0010',  expected: 10
}];

// re-used variable
var count = list.length;
var out = document.getElementById("out");

function toHTMLItem(item) {
  var result = item.expected + " (";
  if(item.after) result += item.after + " < ";
  result += "<b>" + item.id + "</b>";
  if(item.before) result += " &lt; " + item.before;
  result += ")";
  return result;
}

function toHTMLList(list) {
  var result = "<p>";
  for(var i = 0; i < count; ++i) {
    result += toHTMLItem(list[i]) + "<br>";
  }
  result += "</p>";
  return result;
}

// out.innerHTML += toHTMLList(list);

var befores = { };
for(var i = 0; i < count; ++i)
	befores[list[i].id] = null;
function insert(before, after) {
	if(!before || !after)
		return;
	befores[before] = { next:befores[before], id:after };
}
for(var i = 0; i < count; ++i) {
	var item = list[i];
	insert(item.after, item.id);
	insert(item.id, item.before);
}

function toHTMLTable(table, list) {
  var result = "<p>";
  var count = list.length;
  for(var i = 0; i < count; ++i) {
    var id = list[i].id;
    result += id + " < ";
    for(var item = table[id]; item != null; item = item.next) {
      result += item.id + ", ";
    }
    result += "o<br>";
  }
  result += "</p>";
  return result;
}

// out.innerHTML += toHTMLTable(befores, list);

// next build a lookup-table of a completed adjacency list
var lookup = { };
for(var i = 0; i < count; ++i) {
	var id = list[i].id;
	var beforeList = [id];
	var beforeSet = { };
	beforeSet[id] = 1;
	// use "A before B" and "B before C" to gain "A before C"
	for(var j = 0; j < beforeList.length; ++j) {
		for(var item = befores[beforeList[j]]; item != null; item = item.next) {
			if(!beforeSet[item.id]) {
				beforeList.push(item.id);
				beforeSet[item.id] = 1;
			}
		}
	}
	beforeList = beforeList.slice(1, beforeList.length);
	beforeList.sort();
	lookup[id] = beforeSet;
}

function toHTMLLookup(lookup, list) {
  var result = "<p>";
  for(var i = 0, imax = list.length; i < imax; ++i) {
    var id = list[i].id;
    var bs = lookup[id];
    result += id + " < ";
    for(var j = 0, jmax = imax; j < jmax; ++j) {
      if(j == i) continue;
      if(bs[list[j].id]) result += list[j].id + ", ";
    }
    result += "o<br>";
  }
  result += "</p>";
  return result;
}

// out.innerHTML += toHTMLLookup(lookup, list);

// Search function in befores:
// a) if rhs is present in union of befores set lhs < rhs
// b) if rhs is not present in union of befores set rhs < lhs
list.sort((function() {
  var enableAlert = true;
  return function(lhs, rhs) {
    if(!lhs.after || !rhs.before) return -1;
    if(!lhs.before || !rhs.after) return 1;
    if(lhs.id === rhs.id) return 0;
    // different ids guaranteed, doesn't matter if lookup[id] contains id itself
    var result = lookup[lhs.id][rhs.id] ? -1 : 1;
    // expect reversing lhs and rhs to get inverse result
    var expected = lookup[rhs.id][lhs.id] ? 1 : -1;
    if(enableAlert && result != expected) {
      // restrict to a single alert per execution
      enableAlert = false;
      out.innerHTML += "<p><b>ALERT</b> unresolved adjacency between " + lhs.id + " and " + rhs.id + "!</p>";
    }
    return result;
  };
})());

// out.innerHTML += toHTMLList(list);

var error = count;
for(var i = 0; i < error; ++i) {
	if(list[i].expected != i)
		error = i;
}
if(error < count) {
  out.innerHTML += "<h2>error!</h2><p>list[" + error + "] contains " + toHTMLItem(list[i]) + "</p>";
} else {
  out.innerHTML += "<h2>success!</h2>";
}
// Finally print the output
out.innerHTML += toHTMLList(list);

<div id="out"/>

对于此问题的一般主题，请考虑 topological sorting .