栈和队列_leetcode279-经典的层次遍历

#coding=utf-8
#点评：
#递归思想加层次遍历，很经典的解题模板思想  20181220



class Solution(object):
 def numSquares(self, n):
 """
 :type n: int
 :rtype: int
 """

 pair = (n,0)
 queue = []
 queue.append(pair)

 while queue:
 curPair = queue.pop(0)
 num = curPair[0]
 step = curPair[1]

 if num == 0:
 return step

 for i in range(1,num+1):
 if num - i *i >= 0:
 nextPair = (num-i*i,step+1)
 queue.append(nextPair)
 else:
 break

class Solution2(object):
 def numSquares(self, n):
 """
 :type n: int
 :rtype: int
 """
 pair = (n, 0)
 queue = []
 queue.append(pair)

 #matrix = [[0 for i in range(3)] for i in range(3)]
 visit = [0 for i in range(n+1)]
 visit[n] = 1

 while queue:
 curPair = queue.pop(0)
 num = curPair[0]
 step = curPair[1]

 if num == 0:
 return step


 for i in range(1, num + 1):
 nextNum = num - i*i
 if nextNum >= 0:
 if visit[nextNum] == 0:
 nextPair = (nextNum, step + 1)
 queue.append(nextPair)
 visit[nextNum] = 1
 else:
 break


class Solution3(object):
 def numSquares(self, n):
 """
 :type n: int
 :rtype: int
 """
 pair = (n, 0)
 queue = []
 queue.append(pair)

 #matrix = [[0 for i in range(3)] for i in range(3)]
 visit = [0 for i in range(n+1)]
 visit[n] = 1

 while queue:
 curPair = queue.pop(0)
 num = curPair[0]
 step = curPair[1]

 if num == 0:
 return step


 for i in range(1, num + 1):
 nextNum = num - i*i
 if nextNum >= 0:
 if visit[nextNum] == 0:

 if nextNum == 0:
 return step+1

 nextPair = (nextNum, step + 1)
 queue.append(nextPair)
 visit[nextNum] = 1
 else:
 break



# 动态规划
class Solution4(object):
 def numSquares(self, n):

 return self.squareNum(n)

 def squareNum(self, n):

 self.memo = [n for i in range(n + 1)]

 self.memo[0] = 0
 self.memo[1] = 1

 # #memo[2]
 # for i in range(1, 2):
 # if n - i ** 2 >= 0:
 # self.memo[2] = min(self.memo[2],1+self.memo[n - i ** 2])
 #
 #
 # #memo[3]
 # for i in range(1, 3):
 # if n - i ** 2 >= 0:
 # self.memo[3] = min(self.memo[3],1+self.memo[n - i ** 2])



 for i in range(1,n+1):
 for j in range(1,i):
 if i - j ** 2 >= 0:
 self.memo[i] = min(self.memo[i], 1 + self.memo[i - j ** 2])


 return self.memo[n]s = Solution2()print s.numSquares(13)