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LintCode刷题笔记-- Edit distance

描述:

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

  • Insert a character
  • Delete a character
  • Replace a character

解题思路:

这一题做的很快,与之前的字符串匹配问题一样,在word1与word2的两个向量上分解字符串,同时遍历两个字符串:分别在两个子串中进行比较,

对于子串中拥有相同的字符的情况下:加入这个字符与不加入这个字符的意义是相同的,不会有更多的变化,所以有公式:

dp[i][j] = dp[i-1][j-1] 

当两个子串的匹配的字符不同的情况下:有三种情况可以达到当前状态,修改1位,删除1位,加入1位,三个分别的位置为dp[i-1][j], dp[i][j-1],dp[i-1][j-1]

且三个位置记录着之前状态下,所存在的最小变化情况。所以要在当前状态下达到最小,则需要在之前最小的情况下加上1,所以公式有:

dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])+1

参考代码:

 public int minDistance(String word1, String word2) {        // write your code here        int[][] dp = new int[word1.length()+1][word2.length()+1];                for(int i = 0; i<=word1.length(); i++){            dp[i][0] = i;        }                for(int j = 0; j<=word2.length(); j++){            dp[0][j] = j;        }                for(int i=1; i<= word1.length(); i++){            for(int j=1; j<=word2.length(); j++){                if(word1.charAt(i-1)==word2.charAt(j-1)){                    dp[i][j] = dp[i-1][j-1];                }else{                    dp[i][j] = Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1]))+1;                }            }        }                return dp[word1.length()][word2.length()];            }

 

LintCode刷题笔记-- Edit distance