LeetCode:Edit Distance

Edit Distance

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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:

a) Insert a character
b) Delete a character
c) Replace a character

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 (M) One Edit Distance

思路：动规

设：word1="abcd"，word2="bbc"。

则其状态转移过程为：

  Ø a b c d
Ø 0 1 2 3 4
b 1 1 1 2 3
b 2 2 1 2 3
c 3 3 2 1 2

其状态转移方程为：

dp[i][j] = dp[i-1][j-1],if word1[i-1]==word2[j-1];dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1,otherwise;

c++ code:

/*  Ø a b c dØ 0 1 2 3 4b 1 1 1 2 3b 2 2 1 2 3c 3 3 2 1 2*/class Solution {public:    int minDistance(string word1, string word2) {                int m = word1.length();        int n = word2.length();                int dp[m+1][n+1];                for(int i=0;i<=m;i++) dp[i][0] = i; // 初始化第一列        for(int j=0;j<=n;j++) dp[0][j] = j; // 初始化第一行                for(int i=1;i<=m;i++) {            for(int j=1;j<=n;j++) {                                if(word1[i-1]==word2[j-1])                    dp[i][j] = dp[i-1][j-1];                else                    dp[i][j] = min3(dp[i-1][j-1],dp[i-1][j],dp[i][j-1]) + 1;                            }        }        return dp[m][n];    }        // 自定义函数    int min3(int x, int y,int z) {        return min(x,min(y,z));    }};