递归与动态规划---矩阵的最小路径和

【题目】

　　给定一个矩阵ｍ，从左上角开始每次都只能向下或者向右走，最后到达右下角的位置，路径上所有的数字累加起来就是路径和，返回所有路径中最小的路径和。

【基本思路】
　
　　生成和m同样大小的矩阵dp，dp[i][j]的值表示从左上角走到位置(i, j)的最小路径和，矩阵的第一行和第一列的值可以先确定，其他的位置dp[i][j]的值等于min(dp[i-1][j], dp[i][j-1])。

#python3.5#经典动态规划实现def minPathSum1(m):    if m == None or len(m) == 0 or m[0] == None or len(m[0]) == 0:        return 0    dp = [[0 for i in range(len(m[0]))] for j in range(len(m))]    dp[0][0] = m[0][0]    for i in range(1, len(m[0])):        dp[0][i] = dp[0][i-1] + m[0][i]    for j in range(1, len(m)):        dp[j][0] = dp[j-1][0] + m[j][0]    for i in range(1, len(m)):        for j in range(1, len(m[0])):            dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + m[i][j]    return dp[len(m)-1][len(m[0])-1]#经过空间压缩的动态规划def minPathSum2(m):    if m == None or len(m) == 0 or m[0] == None or len(m[0]) == 0:        return 0    more = max(len(m), len(m[0]))    less = min(len(m), len(m[0]))    rowmore = True if more == len(m) else False    dp = [0 for i in range(less)]    dp[0] = m[0][0]    for i in range(1, less):        dp[i] = dp[i-1] + m[0][i] if rowmore else m[i][0]    for i in range(1, more):        dp[0] = dp[0] + m[i][0] if rowmore else m[0][i]        for j in range(1, less):            dp[j] = min(dp[j-1], dp[j]) + m[i][j] if rowmore else m[j][i]    return dp[less-1]