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

思路就是用这三种方式分别去找最小的，递归做，贴一下代码

public int minDistance(String word1, String word2) {
        if (word1.equals(word2)) {
            return 0;
        }
        if (word1.length() < word2.length()) {
            String t = word2;
            word2 = word1;
            word1 = t;
        }
        int pos = 0;
        for (; pos < word2.length(); pos++) {
            if (word1.charAt(pos) != word2.charAt(pos)) {
                break;
            }
        }
        if (pos == word2.length()) {
            return word1.length() - word2.length();
        } else {
            int add = 1 + minDistance(word1.substring(pos), word1.charAt(pos)
                    + word2.substring(pos));
            int delete = 1 + minDistance(
                    pos == word1.length() - 1 ? "" : word1.substring(pos + 1),
                    word2.substring(pos));
            int change = 1 + minDistance(
                    pos == word1.length() - 1 ? "" : word1.substring(pos + 1),
                    pos == word2.length() - 1 ? "" : word2.substring(pos + 1));
            add = add > delete ? delete : add;
            change = change > add ? add : change;
            return change;
        }
    }

然后发现超时了0.0    忽然想起来编程之美上有这个题，查了一下，在第三章，书上的代码和我这差不多啊。。。编程之美上的解法都过不了是么。。。

分析了一下，由于多次递归，时间大部分都消耗在递归上了。

假设word1的长度为i，word2的长度为j

则word1与word2的距离只可能与这三种情况有关：

1.word1.substring(0,i)与word2之间的距离（删除）

2.word1与word2.substring(0,j)之间的距离（添加）

3.word1.substring(0,i)与word2.substring(0,j)之间的距离（更改或不变）

因此，可以将刚才自底向上的解法转化为自顶向下去计算，这样就消除了递归

最初设计的是word1.length()-1*word2.length()-1的矩阵grid，但是在矩阵grid[0][0]赋值的时候遇到了问题，因为最短距离时，word1[0]与word2[0]不一定对应，因此，多加一行一列，grid[i][j]表示word1的(0,i)子串与word2的(0,j)子串的最小距离，如下表所示：

以word1=sea，word2=eat为例：

“” “”	“” “e”	"" "ea"	"" "eat"
"s" ""	"s" "e"	"s" "ea"	"s" "eat"
"se" ""	"se" "e"	"se" "ea"	"se" "eat"
"sea" ""	"sea" "e"	"sea" "ea"	"sea" "eat"

 

 

 

 

 

 

 

 

 

 

gird[i][j] = min(grid[i-1][j]+1,grid[i][j-1]+1,gird[i-1][j-1]+(word1.charAt(i)==word2.charAt(j)?0:1));

第0行和第0列要判断边界条件，初始化grid[0][0]=0;

结果就是矩阵grid右下角元素的值，代码如下：

public int minDistance(String word1, String word2) {
        if(word1.equals("")){
            return word2.length();
        }
        if(word2.equals("")){
            return word1.length();
        }
        int grid[][] = new int[word1.length()+1][word2.length()+1];
        for (int i = 0; i <= word1.length(); i++) {
            for (int j = 0; j <= word2.length(); j++) {
                if (i == 0 && j == 0) {
                    grid[i][j] = 0;                        
                } else if (i == 0) {
                    grid[i][j] = grid[i][j - 1] + 1;
                } else if (j == 0) {
                    grid[i][j] = grid[i - 1][j] + 1;
                } else {
                    grid[i][j] = grid[i - 1][j] > grid[i][j - 1] ? grid[i][j - 1] + 1
                            : grid[i - 1][j] + 1;
                    int count = 0;
                    if(word1.charAt(i-1)!=word2.charAt(j-1)){
                        count = 1;
                    }
                    grid[i][j] = grid[i][j]<grid[i-1][j-1]+count?grid[i][j]:grid[i-1][j-1]+count;
                }
            }
        }
        return grid[word1.length()][word2.length()];
    }