如何使用SequenceEqual比较坐标?

时间:2013-05-21 01:24:25

标签: c# list comparison

如果嵌套的int[]包含坐标x,y,我如何使用SequenceEqual比较它们?

列表是一组坐标。我想检查每个其他List以查看它们是否具有相同数量的坐标以及相同的坐标值。如果它们都匹配,我想删除多余的。否则,留下它。

  private List<List<int[]>> combineList(List<List<int[]>> matches){
        Debug.Log (matches.Count());

        foreach(List<int[]> tileGroup in matches){
            foreach(List<int[]> other in matches){
                if(other == tileGroup) continue;

                if(sequenceEqual(tileGroup, other)){
                    matches.Remove(other);
                }   


            }
        }

        Debug.Log (matches.Count());

        return matches;
    }

    private bool sequenceEqual(List<int[]> groupA, List<int[]> groupB){
        if(groupA.Count() == groupB.Count()){
            int i = 0, j = 0;
            Dictionary<int, int[]>  dictA = new Dictionary<int, int[]>(), 
                                    dictB = new Dictionary<int, int[]>();   

            foreach(int[] coordinate in groupA){
                dictA.Add (i, coordinate);
                i++;
            }

            foreach(int[] coordinate in groupB){
                dictB.Add (j, coordinate);  
                j++;
            }

            return dictA.Values.SequenceEqual(dictB.Values);
        } 

        return false;
    }

1 个答案:

答案 0 :(得分:1)

可能最快的方法是实施IEqualityComparer<int[]>

class IntArrayEqualityComparer : IEqualityComparer<int[]>
{
    public bool Equals(int[] x, int[] y)
    {
        if (ReferenceEquals(x, y)) return true;
        if (ReferenceEquals(null, x)) return false;
        if (ReferenceEquals(null, y)) return false;
        if (x.Length != y.Length) return false;
        for (var i = 0; i < x.Length; i++)
        {
            if (x[i] != y[i]) return false;
        }
        return true;
    }

    public int GetHashCode(int[] x)
    {
        if (x == null) return 0;
        var hashCode = 0;
        for (var i = 0; i < x.Length; i++)
        {
            hashCode = (32 * hashCode) + x[i];
        }
        return hashCode;
    }
}

然后使用重载版本的IEnumerable<TSource>.SequenceEqual

private bool sequenceEqual(List<int[]> groupA, List<int[]> groupB)
{
    if (ReferenceEquals(groupA, groupB)) return true;
    if (ReferenceEquals(null, groupA)) return false;
    if (ReferenceEquals(null, groupB)) return false;
    return groupA.SequenceEqual(groupB, new IntArrayEqualityComparer());
}

从长远来看,创建一个简单实现Coordinates的{​​{1}}类型可能会有所帮助,在这种情况下,您只需比较两个IEquatable<Coordinates>个对象。