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583. 两个字符串的删除操作

方法：dp

状态表示：以i-1和j-1为结尾的字符串world1和world2，抵达相同的字符串所需的最少操作数

属性：最小值

状态计算：world1[i-1]和world2[j-1]相同dp[i][j] = dp[i-1][j-1];

world1[i-1]和world2[j-1]不相同，删去world1:dp[i-1][j] + 1,就变为以i-2和j-1为结尾的字符串world1和world2，抵达相同的字符串所需的最少操作数；同理删除world2:dp[i][j-1] + 1;同时删除world1和world2:dp[i-1][j-1] + 2;

细心的话可以发现dp[i-1][j] + 1 = dp[i-1][j-1] = dp[i][j-1] + 1

所以递推公式dp[i][j] = min(dp[i-1][j] + 1, dp[i][j-1] + 1)

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 = 0; i <= n; ++i) dp[i][0] = i;for (int i = 0; i <= m; ++i) dp[0][i] = i;for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++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, dp[i][j-1] + 1);}return dp[n][m];}
};

$时间复杂度O(n*m)，空间复杂度O(n*m);

方法2：dp

状态表示：以i-1和j-1为结尾的字符串world1和world2，最大的相同子序列的集合为dp[i][j]

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 <= n; ++i)for (int j = 1; j <= m; ++j) {if (word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1] + 1;else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);}return n + m - dp[n][m] * 2;}
};

$时间复杂度O(n*m)，空间复杂度O(n*m);

72. 编辑距离

方法：dp

简单说一下增加和删除的效果是一样的所以就统一删除了

替换就是在dp[i-1][j-1]的基础上加一个操作

其他的都差不多

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 = 0; i <= n; ++i) dp[i][0] = i;for (int i = 0; i <= m; ++i) dp[0][i] = i;for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++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], dp[i-1][j-1]}) + 1;}return dp[n][m];}
};

$时间复杂度O(n*m)，空间复杂度O(n*m);