kth sum of two sorted arrays

转自：http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs;action=display;num=1132204952;start=50

 

一个解法可以是

q : priority queue (decreasing) := empty priority queue
add (0, 0) to q with priority a[0] + b[0]
while k > 0:
    k--
    x := pop q
    output x
    (i, j) : tuple of int,int := position of x
    if i < m:
        add (i + 1, j) to q with priority a[i + 1] + b[j]
    if j < n:
        add (i, j + 1) to q with priority a[i] + b[j + 1]

  
  
  
  

   
   
   
   
The loop is executed k times. 
    
 
     
There is one pop operation per iteration. 
 
     
There are up to two insert operations per iteration. 
 
    
   
   
   
   
The maximum size of the priority queue is O(min(m, n)) O(m + n). 
   
   
   
   
The priority queue can be implemented with a binary heap giving log(size) pop and insert. 
   
   
   
   
Therefore this algorithm is O(k * log(min(m, n))) O(k * log(m + n)). 
  
  
  
  

但注意可能有重复的值被添加，所以在填入q中时要先判断是否已存在，出队时删除其存在。用一个set可以方便的判断。