Leetcode:72. Edit Distance
Description
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
- 动态规划,dp[i][j]表示words1[0..i]和words2[0..j]最小的编辑距离是多少
- dp[i][j] =min( min(dp[i - 1][j - 1] + count, dp[i - 1][j] + 1), dp[i][j - 1] + 1)
- 其中当words1[i]==words2[j]时,count = 1; 否则,count = 0;
代码
class Solution {
public:
int minDistance(string word1, string word2) {
int len1 = word1.size(), len2 = word2.size();
if (len1 == 0) return len2;
if (len2 == 0) return len1;
vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0));
dp[0][0] = 0;
for (int i = 1; i <= len1; ++i)
dp[i][0] = i;
for (int j = 1; j <= len2; ++j)
dp[0][j] = j;
for (int i = 1; i <= len1; ++i){
for (int j = 1; j <= len2; ++j){
int count = (word1[i - 1] == word2[j - 1] ? 0 : 1);
dp[i][j] = min(min(dp[i - 1][j - 1] + count, dp[i - 1][j] + 1), dp[i][j - 1] + 1) ;
}
}
return dp[len1][len2];
}
};