动态规划_leetcode279(经典栈板)

标签：tco   obj   type   font   utf-8   lse   mat   step   code   

#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


s = Solution2()

print s.numSquares(13)

动态规划_leetcode279(经典栈板)

标签：tco   obj   type   font   utf-8   lse   mat   step   code   

原文地址：https://www.cnblogs.com/lux-ace/p/10546545.html