[LeetCode] Edit Distance 编辑距离
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
这道题让求从一个字符串转变到另一个字符串需要的变换步骤,共有三种变换方式,插入一个字符,删除一个字符,和替换一个字符。根据以往的经验,对于字符串相关的题目十有八九都是用动态规划Dynamic Programming来解,这道题也不例外。这道题我们需要维护一个二维的数组dp,其中dp[i][j]表示从word1的前i个字符转换到word2的前j个字符所需要的步骤。那我们可以先给这个二维数组dp的第一行第一列赋值,这个很简单,因为第一行和第一列对应的总有一个字符串是空串,于是转换步骤完全是另一个字符串的长度。跟以往的DP题目类似,难点还是在于找出递推式,我们可以举个例子来看,比如word1是“bbc",word2是”abcd“,那么我们可以得到dp数组如下:
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
我们通过观察可以发现,当word1[i] == word2[j]时,dp[i][j] = dp[i - 1][j - 1],其他情况时,dp[i][j]是其左,左上,上的三个值中的最小值加1,那么可以得到递推式为:
dp[i][j] = / dp[i - 1][j - 1] if word1[i - 1] == word2[j - 1]
\ min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1 else
复制代码
class Solution {
public:
int minDistance(string word1, string word2) {
int n1 = word1.size(), n2 = word2.size();
int dp[n1 + 1][n2 + 1];
for (int i = 0; i <= n1; ++i) dp[i][0] = i;
for (int i = 0; i <= n2; ++i) dp[0][i] = i;
for (int i = 1; i <= n1; ++i) {
for (int j = 1; j <= n2; ++j) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
} else {
对i替换变为j; i删除变为j; i增加变为j;
dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
}
}
}
return dp[n1][n2];
}
};
复制代码