动态规划_leetcode62

最新推荐文章于 2025-08-14 22:07:01 发布

weixin_30321709

最新推荐文章于 2025-08-14 22:07:01 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/lux-ace/p/10546252.html

此博客为转载内容，转载自https://www.cnblogs.com/lux-ace/p/10546252.html ，涉及数据结构与算法领域。

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#coding=utf-8

# 递归


# 深度优先遍历，找全路径     20190306 找工作期间
class Solution1(object):
def uniquePaths(self, m, n):
"""
        :type m: int
        :type n: int
        :rtype: int
        """

        if m == 0 and n == 0:
return

        self.m = m
self.n = n
self.res = 0

        self.direction = [[0,1],[1,0]]

self.visit = [[False for i in range(n)] for i in range(m)]

self.findPath(0,0)

print self.res

return self.res


def inArea(self,x,y):
return x >= 0 and x < self.m and y >= 0 and y < self.n

def findPath(self,x,y):

if x == self.m-1 and y == self.n-1:
self.res += 1
            return



        self.visit[x][y] = True

        for item in self.direction:
            newX = x + item[0]
            newY = y + item[1]

if self.inArea(newX,newY) and not self.visit[newX][newY]:
self.findPath(newX,newY)

self.visit[x][y] = False




# s = Solution1()
#
# s.uniquePaths(23,12)



# 记忆化递归
class Solution2(object):
def uniquePaths(self, m, n):
"""
        :type m: int
        :type n: int
        :rtype: int
        """

        if m == 0 and n == 0:
return

        self.m = m
self.n = n

self.res = 0

        # memo[x][y] 矩阵的方案数
        # 矩阵坐标[x][y] 到 矩阵[1][1]的 路径数
        self.memo = [[-1 for i in range(n+1)] for i in range(m+1)]

for i in range(n+1):
self.memo[0][i] = 0

        for i in range(m+1):
self.memo[i][0] = 0

        self.memo[1][1] = 1

        #self.visit = [[False for i in range(n)] for i in range(m)]

        return self.findPath(m, n)

def findPath(self, x, y):

if x == 1 and y == 1:
return 1

        if x == 0 or y == 0:            return 0        #        # self.memo[x][y] = self.memo[x-1][y] + self.memo[x][y-1]        if self.memo[x][y] == -1:            if self.memo[x-1][y] != -1:                self.memo[x][y] = self.memo[x-1][y]            elif  self.memo[x-1][y] == -1:                self.memo[x-1][y] = self.findPath(x-1,y)                self.memo[x][y] = self.memo[x - 1][y]            if self.memo[x][y-1] != -1:                self.memo[x][y] += self.memo[x][y-1]            elif self.memo[x][y-1] == -1:                self.memo[x][y-1] = self.findPath(x,y-1)                self.memo[x][y] += self.memo[x][y-1]        return self.memo[x][y]# s = Solution2()## print s.uniquePaths(23,12)# 动态规划class Solution3(object):    def uniquePaths(self, m, n):        """        :type m: int        :type n: int        :rtype: int        """        if m == 0 and n == 0:            return        return self.findPath(m, n)    def findPath(self, m, n):        # memo[x][y] 矩阵的方案数        memo = [[-1 for i in range(n + 1)] for i in range(m + 1)]        for i in range(n + 1):            memo[0][i] = 0        for i in range(m + 1):            memo[i][0] = 0        memo[1][1] = 1        for i in range(1,m+1):            for j in range(1,n+1):                if i == 1 and j == 1:                    continue                memo[i][j] = memo[i-1][j] +memo[i][j-1]        return memo[m][n]s = Solution3()print s.uniquePaths(7,3)