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583. 两个字符串的删除操作

题目链接

思路

1. 确定dp数组（dp table）以及下标的含义
   dp[i][j]：以i-1为结尾的字符串word1，和以j-1位结尾的字符串word2，想要达到相等，所需要删除元素的最少次数

2. 确定递推公式
   当word1[i - 1]=word2[j - 1]，dp[i][j] = dp[i - 1][j - 1];
   当word1[i - 1] !=word2[j - 1]：

   • 情况一：删word1[i - 1]，最少操作次数为dp[i - 1][j] + 1
   • 情况二：删word2[j - 1]，最少操作次数为dp[i][j - 1] + 1
   • 情况三：同时删word1[i - 1]和word2[j - 1]，操作的最少次数为dp[i - 1][j - 1] + 2，它包含在情况1和2里面

   取最小值，则递推公式可简化为：dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);

3. dp数组初始化
   自行理解
   vector<vector> dp(word1.size() + 1, vector(word2.size() + 1));
   for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
   for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;

4. 确定遍历顺序
   递推公式可以看出从上到下，从左到右

5. 举例推导dp数组

代码实现

class Solution {
public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word1.size()+1,vector<int>(word2.size()+1,0));for(int i=1;i<=word1.size();i++)dp[i][0]=i;for(int j=1;j<=word2.size();j++)dp[0][j]=j;for(int i=1;i<=word1.size();i++){for(int j=1;j<=word2.size();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[word1.size()][word2.size()];}
};

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72. 编辑距离

题目链接

思路

和上面那道题几乎一模一样
只不过在编辑时不仅有增删操作，还有替换操作，所以
if(word1[i-1]!=word2[j-1])dp[i][j]=min(dp[i-1][j]+1,min(dp[i][j-1]+1,dp[i-1][j-1]+1));
其中：
增删：dp[i-1][j]+1，dp[i][j-1]+1
替换：dp[i-1][j-1]+1

代码实现

class Solution {
public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word1.size()+1,vector<int>(word2.size()+1,0));for(int i=1;i<=word1.size();i++)dp[i][0]=i;for(int j=1;j<=word2.size();j++)dp[0][j]=j;for(int i=1;i<=word1.size();i++){for(int j=1;j<=word2.size();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][j-1]+1,dp[i-1][j-1]+1));}}return dp[word1.size()][word2.size()];}
};

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