- 分析
dp[i][j]表示word1的前i个字符和word2的前j个字符编程相同的字符串所需要的最小的操作数。状态转移方程见代码。
- 代码
class Solution {
public:
int minDistance(string word1, string word2) {
int len1 = word1.length();
int len2 = word2.length();
vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0));
for(int i = 0; i <= len1; i++) dp[i][0] = i;
for(int i = 0; i <= len2; i++) dp[0][i] = i;
for(int i = 1; i <= len1; i++){
for(int j = 1; j <= len2; 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], dp[i][j - 1]) + 1;
}
}
}
return dp[len1][len2];
}
};