LeetCode | 72. Edit Distance

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

思路

对于字符串 str1 和 str2，dp[i][j]为考虑str1的0~i, str2的0~j的最短编辑距离。

• 若str1[i] == str2[j]：dp[i][j] = dp[i-1][j-1]
• 若str1[i] != str2[j]，有三种操作，让两个字符串对齐：
  
  • 1、替换掉这个字符，使得str1[i] == str2[j]，dp[i][j] = dp[i-1][j-1]+1
  • 2、删去str1中的这个字符，使得问题变为考虑str1[0~i-1]和str2[0~j]对齐问题,dp[i][j] = dp[i-1][j]+1
  • 3、删去str2[j]，dp[i][j] = dp[i][j-1]+1
    综合上面三种情况，找出最小值作为dp[i][j]的值

难点在于边界情况的处理！

代码

代码1：静态数组，内存开销大，86 ms

class Solution {public:    int mymin(int a,int b,int c)    {        return min(min(a,b),c);    }    int minDistance(string word1, string word2)    {        int len1 = word1.length(), len2 = word2.length();        int dp[500][500] = {};        int flag1 = 1, flag2 = 1;        if(len1 == 0)            return len2;        if(len2 == 0)            return len1;        if(word1[0]==word2[0])        {            dp[0][0] = 0;            flag1 = flag2 = 0;        }        else            dp[0][0] = 1;        //1-1:dinitrophenylhydrazine        //1-2:phenylhydrazine        //2-1:sea        //2-2:eat        for(int i=1;i<len1;i++)        {            /*解释下面这个 if             此后只要删除无关字符即可。例如：abcd  caret             dp[0][1]=1;             dp[0][2]=2，而不是3             */            if(word1[i] == word2[0])            {                dp[i][0] = i;                flag1 = 0;            }            else                dp[i][0] = i+flag1;        }        for(int j=1;j<len2;j++)        {            if(word1[0] == word2[j])            {                dp[0][j] = j;                flag2 = 0;            }            else                dp[0][j] = j+flag2;        }        for(int i=1;i<len1;i++)        {            for(int j=1;j<len2;j++)            {                if(word1[i] == word2[j])                    dp[i][j] = dp[i-1][j-1];                else                {                    dp[i][j] = mymin(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1;                }            }        }        return dp[len1-1][len2-1];    }};

代码2：动态数组法，节省内存空间，13 ms

class Solution {public:    int mymin(int a,int b,int c)    {        return min(min(a,b),c);    }    int minDistance(string word1, string word2)    {        int len1 = word1.length(), len2 = word2.length();        vector<vector<int> > dp;        int flag1 = 1, flag2 = 1;        if(len1 == 0)            return len2;        if(len2 == 0)            return len1;        vector<int> tmp;        tmp.push_back(0);        dp.push_back(tmp);        if(word1[0]==word2[0])        {            dp[0][0] = 0;            flag1 = flag2 = 0;        }        else            dp[0][0] = 1;        //1-1:dinitrophenylhydrazine        //1-2:phenylhydrazine        //2-1:sea        //2-2:eat        for(int i=1;i<len1;i++)        {            /*解释下面这个 if             此后只要删除无关字符即可。例如：abcd  caret             dp[0][1]=1;             dp[0][2]=2，而不是3             */            vector<int> t;            if(word1[i] == word2[0])            {                t.push_back(i);                flag1 = 0;            }            else                t.push_back(i+flag1);            dp.push_back(t);        }        for(int j=1;j<len2;j++)        {            if(word1[0] == word2[j])            {                dp[0].push_back(j);                flag2 = 0;            }            else                dp[0].push_back(j+flag2);        }        for(int i=1;i<len1;i++)        {            for(int j=1;j<len2;j++)            {                if(word1[i] == word2[j])                    dp[i].push_back(dp[i-1][j-1]);                else                {                    dp[i].push_back(mymin(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1);                }            }        }        return dp[len1-1][len2-1];    }};

代码3：对边界的处理更熟练，13 ms

class Solution { public:    int minDistance(string word1, string word2) {         int m = word1.length(), n = word2.length();        vector<vector<int> > dp(m + 1, vector<int> (n + 1, 0));        for (int i = 1; i <= m; i++)            dp[i][0] = i;        for (int j = 1; j <= n; j++)            dp[0][j] = j;          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] = min(dp[i - 1][j - 1] + 1, min(dp[i][j - 1] + 1, dp[i - 1][j] + 1));            }        }        return dp[m][n];    }};