Edit Distance

2016-03-15

问题描述

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

思路

将word1转化为目标串word2，题中给了我们三种操作，分别是：

插入一个字符
删除一个字符
替换一个字符

这是一道典型的动态规划（DP）问题，我们可以设，i，j分别是word1和word2的下标。为了后边后续的表述以及代码的可读性，我们的下标以1为开始。

定义dp[i][j] 表示word1[1…i]这个子串转换为word2[1…j]这个子串所需要的最少操作数。

从而我们可以知道：

dp[0][j] = j;
dp[i][0] = i;
if (word1[i] == word2[j]) dp[i][j] = dp[i-1][j-1];
if (word1[i] != word2[j]) dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1;

其中min中的，dp[i-1][j-1]对应替换一个字符，dp[i][j-1]对应插入一个字符，dp[i-1][j]对应删除一个字符。

Code

int minDistance(string word1, string word2) {
    const int n1 = word1.size();
    const int n2 = word2.size();
    if (n1 == 0 || n2 == 0) return abs(n1-n2);
    int **dp = new int*[n1+1];
    for (int k = 0; k <= n1; k++){
        dp[k] = new int[n2+1];
    }
    int i, j;
    //初始化
    for (i = 0; i <= n1; i++){
        dp[i][0] = i;
    }
    for (j = 0; j <= n2; j++){
        dp[0][j] = j;
    }
    for (i = 1; i <= n1; i++){
        for (j = 1; j <= n2; j++){
            if (word1[i-1] == word2[j-1]){
                dp[i][j] = dp[i-1][j-1];
            }
            else{
                dp[i][j] = min( min(dp[i-1][j-1], dp[i][j-1]),dp[i-1][j]) + 1;
            }
        }
    }
    return dp[n1][n2];
}