三种操作的转移:
class Solution {
public:
int minDistance(string word1, string word2) {
int n = word1.size(), m = word2.size();
vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
for(int i = 1; i <= m; i++)
dp[0][i] = i;
for(int i = 1; i <= n; i++)
dp[i][0] = i;
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= m; j++)
{
dp[i][j] = dp[i - 1][j - 1] + (word1[i - 1] != word2[j -1]);//替换一个 这里word1 word2的是index所以减1
dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);//删掉一个
dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1);//添加一个
}
}
return dp[n][m];
}
};
第二次写题
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> distance = vector<vector<int>>(word1.size() + 1, vector<int>(word2.size() + 1, INT_MIN));
for(int i = 0; i <= word1.size(); i++)
distance[i][0] = i;
for(int i = 0; i <= word2.size(); i++)
distance[0][i] = i;
for(int i = 1; i <= word1.size(); i++)
{
for(int j = 1; j <= word2.size(); j++)
{
distance[i][j] = distance[i - 1][j - 1] + (word1[i - 1] != word2[j - 1]); //替换;
distance[i][j] = min(distance[i][j], distance[i - 1][j] + 1);//插入
distance[i][j] = min(distance[i][j], distance[i][j - 1] + 1);//删除
}
}
return distance[word1.size()][word2.size()];
}
};
第二次写题
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> distance = vector<vector<int>>(word1.size() + 1, vector<int>(word2.size() + 1, INT_MIN));
for(int i = 0; i <= word1.size(); i++)
distance[i][0] = i;
for(int i = 0; i <= word2.size(); i++)
distance[0][i] = i;
for(int i = 1; i <= word1.size(); i++)
{
for(int j = 1; j <= word2.size(); j++)
{
distance[i][j] = distance[i - 1][j - 1] + (word1[i - 1] != word2[j - 1]); //替换;
distance[i][j] = min(distance[i][j], distance[i - 1][j] + 1);//插入
distance[i][j] = min(distance[i][j], distance[i][j - 1] + 1);//删除
}
}
return distance[word1.size()][word2.size()];
}
};