我正在在线解决一个难题,偶然发现了这个问题,给定一个2D矩阵和一个数字k,我需要返回矩阵中第k个最小的元素。
matrix = [
[ 1, 5, 9],
[10, 11, 13],
[12, 13, 15]
],
k = 8,
return 13.
我可以通过自己的二进制堆实现解决此问题。由于我正在准备面试,因此不确定在所有情况下实施自己的堆是否可以接受。所以我试图用SortedList / SortedSet解决这个问题。
我基本上是在创建一个Point对象,该对象接受索引i,j和i,j处的值。我已经实现了IComparable和IEquatable。但是由于某些原因,在上面的示例中,索引1,2(值13)的Point对象和索引2,1(值13)的Point对象在不应该相等时被视为相等。使用SortedList时,我得到一个ArgumentException,同时SortedSet只是覆盖了现有对象。我的IEquatable,IComparable实现是否错误?我仔细检查了它们是否生成了不同的哈希码。
P.S。这不是作业问题。我正在通过在线面试准备平台解决问题。
public class Solution {
public int KthSmallest(int[,] matrix, int k) {
int rows = matrix.GetLength(0);
int cols = matrix.GetLength(1);
SortedSet<Point> pq = new SortedSet<Point>();
bool[,] visited = new bool[rows, cols];
int count = 1;
int i=0, j=0;
var start = new Point(i, j, matrix[i, j]);
pq.Add(start);
visited[0, 0] = true;
while(true) {
k--;
var curr = pq.Min;
pq.Remove(pq.First());
if(k == 0)
return curr.val;
i = curr.x + 1;
j = curr.y;
if(i < rows && j < cols && !visited[i, j]) {
var next = new Point(i, j, matrix[i, j]);
Console.WriteLine(next.GetHashCode());
Console.WriteLine(i+", "+j+", "+next.val);
pq.Add(next);
visited[i, j] = true;
}
i = curr.x;
j = curr.y + 1;
if(i < rows && j < cols && !visited[i, j]) {
var next = new Point(i, j, matrix[i, j]);
Console.WriteLine(next.GetHashCode());
Console.WriteLine(i+", "+j+", "+next.val);
pq.Add(next);
visited[i, j] = true;
}
}
}
}
public class Point : IComparable<Point>, IEquatable<Point>
{
public int x { get; private set; }
public int y { get; private set; }
public int val { get; private set; }
public Point(int x, int y, int val)
{
this.x = x;
this.y = y;
this.val = val;
}
public int CompareTo(Point that)
{
if(this.val == that.val) {
if(this.x == that.x) {
return this.y.CompareTo(that.y);
}
else {
this.x.CompareTo(that.x);
}
}
return val.CompareTo(that.val);
}
public override bool Equals(object obj)
{
if (ReferenceEquals(null, obj)) return false;
if (ReferenceEquals(this, obj)) return true;
if (obj.GetType() != this.GetType()) return false;
return Equals((Point)obj);
}
public bool Equals(Point p1) {
return x == p1.x && y == p1.y && val == p1.val;
}
public override int GetHashCode() {
long hashCode = ((x * 31 + y) * 31 + val);
return hashCode.GetHashCode();
}
}
答案 0 :(得分:1)
您在CompareTo中缺少return语句。我在下面评论了您的原始照片,以及更正的版本。在比较[2,1]和[1,2]时,x值不匹配,但是当您点击this.x.CompareTo时,您实际上不会返回该比较,因此将返回值比较。
您有:
public int CompareTo(Point that)
{
if(this.val == that.val) {
if(this.x == that.x) {
return this.y.CompareTo(that.y);
}
else {
//****MISSING RETURN STATEMENT -
//will return the val.ComapreTo statement after
//it leaves this block*****
this.x.CompareTo(that.x);
}
}
return val.CompareTo(that.val);
}
您需要:
public int CompareTo(Point that)
{
if(this.val == that.val) {
if(this.x == that.x) {
return this.y.CompareTo(that.y);
}
else {
return this.x.CompareTo(that.x);
}
}
return val.CompareTo(that.val);
}