day55

1、leetcode583 两个字符串的删除操作

1. 动归五步法

   1. 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. 递推公式
      1. word1[i-1] == word2[j-1]
         • dp[i] [j] = dp[i-1] [j-1]
      2. word1[i-1] != word2[j-1]
         1. 删word1[i - 1]，dp[i] [j] = dp[i-1] [j] + 1
         2. 删word2[j - 1]，dp[i] [j] = dp[i] [j-1] + 1
         3. 同时删word1[i - 1]和word2[j - 1]，操作的最少次数为dp[i - 1] [j - 1] + 2
         4. dp[i] [j] = min({dp[i - 1] [j - 1] + 2, dp[i - 1] [j] + 1, dp[i] [j - 1] + 1});
   3. 初始化
      1. dp[i] [0]：word2为空字符串，以i-1为结尾的字符串word1要删除多少个元素，才能和word2相同
         • dp[i] [0] = i
      2. dp[0] [j] = j
   4. 遍历顺序
      • 从上到下，从左到右
   5. 举例

2. 代码

   class Solution {
       public int minDistance(String word1, String word2) {
           int[][] dp = new int[word1.length()+1][word2.length()+1];
   
           for(int i=0; i<=word1.length(); i++) {
               dp[i][0] = i;
           }
           for(int j=1; j<=word2.length(); j++) {
               dp[0][j] = j;
           }
   
           for(int i=1; i<=word1.length(); i++) {
               for(int j=1; j<=word2.length(); j++) {
                   if(word1.charAt(i - 1) == word2.charAt(j-1)) {
                       dp[i][j] = dp[i-1][j-1];
                   } else {
                       dp[i][j] = Math.min(dp[i - 1][j - 1] + 2,
                                           Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));
                   }
               }
           }
   
           return dp[word1.length()][word2.length()];
   
   
       }
   }
   

2、leetcode72 编辑距离

1. 动归五步法

   1. 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，所需要操作的最少次数。
   2. 递推公式
      1. word1[i-1] == word2[j-1]
         • dp[i] [j] = dp[i-1] [j-1]
      2. word1[i-1] != word2[j-1]
         1. 删除
            • 删word1[i - 1]，dp[i] [j] = dp[i-1] [j] + 1【相当于word2 添加一个字符】
            • 删word2[j - 1]，dp[i] [j] = dp[i] [j-1] + 1【相当于word1 添加一个字符】
         2. 插入
         3. 替换
            • word1替换word1[i - 1]，使其与word2[j - 1]相同，此时不用增删加元素。 dp[i] [j] = dp[i - 1] [j - 1] + 1;
      3. dp[i] [j] = min({dp[i - 1] [j - 1], dp[i - 1] [j], dp[i] [j - 1]}) + 1;
   3. 初始化
      1. dp[i] [0]：word2为空字符串，以i-1为结尾的字符串word1要删除多少个元素，才能和word2相同
         • dp[i] [0] = i
      2. dp[0] [j] = j
   4. 遍历顺序
      • 从上到下，从左到右
   5. 举例

2. 代码

   class Solution {
       public int minDistance(String word1, String word2) {
           int[][] dp = new int[word1.length()+1][word2.length()+1];
   
           for(int i=0; i<=word1.length(); i++) {
               dp[i][0] = i;
           }
           for(int j=1; j<=word2.length(); j++) {
               dp[0][j] = j;
           }
   
           for(int i=1; i<=word1.length(); i++) {
               for(int j=1; j<=word2.length(); j++) {
                   if(word1.charAt(i - 1) == word2.charAt(j-1)) {
                       dp[i][j] = dp[i-1][j-1];
                   } else {
                       dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
                   }
               }
           }
   
           return dp[word1.length()][word2.length()];
   
           
       }
   }