題目:
Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:
Insert a character
Delete a character
Replace a character
例子
Example 1:
Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')
問題解析:
求兩個字符串的最小編輯距離。
鏈接:
思路標籤:
算法:動態規劃
解答:
- 使用dp[i][j]用來表示字符串1的0~i-1、字符串2的0~j-1的最小編輯距離;
- 我們可以知道邊界情況:dp[i][0] = i、dp[0][j]=j;
- 同時對於兩個字符串的子串,都能分爲最後一個字符相等或者不等的情況:
- 如果
words[i-1] == words[j-1]
:dp[i][j] = dp[i-1][j-1];也就是說當前的編輯距離和位置i和j的字符無關; - 如果
words[i-1] != words[j-1]
:則存在三種可能的操作:
- 向word1插入:dp[i][j] = dp[i][j-1] + 1;
- 從word1刪除:dp[i][j] = dp[i-1][j] + 1;
- 替換word1元素:dp[i][j] = dp[i-1][j-1] + 1;
class Solution {
public:
int minDistance(string word1, string word2) {
int rows = word1.length();
int cols = word2.length();
vector<vector<int> > dp(rows+1, vector<int>(cols+1, 0));
for(int i=1; i<=rows; ++i)
dp[i][0] = i;
for(int j=1; j<=cols; ++j)
dp[0][j] = j;
for(int i=1; i<=rows; ++i){
for(int j=1; j<=cols; ++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], min(dp[i-1][j], dp[i][j-1])) + 1;
}
}
return dp[rows][cols];
}
};