LeetCode编辑距离

LeetCode编辑距离

原题链接地址

• 解题说明

  本题是一道非常典型的动态规划题

  只有三种操作，代码里面都有描述，非常清晰

  1. 插入一个字符
  2. 删除一个字符
  3. 替换一个字符

• python代码

  class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        #典型的动态规划
        dp = [ [-1 for i in range(len(word2) + 1)] for i in range(len(word1) + 1)]
  
        dp[0][0] = 0
  
        for i in range(len(dp)):
            for j in range(len(dp[0])):
                #插入操作
                if j > 0:
                    if dp[i][j] == -1:
                        dp[i][j] = dp[i][j-1] + 1
                    else: dp[i][j] = min(dp[i][j], dp[i][j-1]+1)
  
                #删除操作
                if  i > 0:
                    if dp[i][j] == -1:
                        dp[i][j] = dp[i-1][j] + 1
                    else: dp[i][j] = min(dp[i][j], dp[i-1][j]+1)
  
                #修改操作，有点特殊，可能不需要修改
                if i > 0 and j > 0:
                    if dp[i][j] == -1:
                        if word1[i-1] == word2[j-1]:
                            dp[i][j] = dp[i-1][j-1]
                        else:
                            dp[i][j] = dp[i-1][j-1] + 1
                    else:
                        if word1[i-1] == word2[j-1]:
                            dp[i][j] = min(dp[i-1][j-1], dp[i][j])
                        else:
                            dp[i][j] = min(dp[i-1][j-1]+1, dp[i][j])
        #print dp
  
        return dp[len(word1)][len(word2)]