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代码随想录算法训练营第五十六天| LeetCode583. 两个字符串的删除操作 72. 编辑距离

代码随想录算法训练营第五十六天| LeetCode583. 两个字符串的删除操作 72. 编辑距离

583. 两个字符串的删除操作

题目:力扣

  1. class Solution {
  2. public:
  3. int minDistance(string word1, string word2) {
  4. vector<vector<int> > dp(word1.size()+1,vector<int>(word2.size()+1));
  5. for(int i = 0 ; i <= word1.size(); ++i) dp[i][0] = i;
  6. for(int j = 0 ; j <= word2.size(); ++j) dp[0][j] = j;
  7. for(int i = 1; i <= word1.size(); ++i){
  8. for(int j = 1; j <= word2.size(); ++j){
  9. if(word1[i-1] == word2[j-1]){
  10. dp[i][j] = dp[i-1][j-1];
  11. }else{
  12. dp[i][j] = min(dp[i][j-1] + 1, dp[i-1][j] + 1);
  13. }
  14. }
  15. }
  16. return dp[word1.size()][word2.size()];
  17. }
  18. };

72. 编辑距离

题目:力扣

  1. class Solution {
  2. public:
  3. int minDistance(string word1, string word2) {
  4. vector<vector<int> > dp(word1.size()+1, vector<int> (word2.size()+1));
  5. for(int i = 0; i <= word1.size(); ++i) dp[i][0] = i;
  6. for(int j = 0; j <= word2.size(); ++j) dp[0][j] = j;
  7. for(int i = 1; i <= word1.size(); ++i){
  8. for(int j = 1; j <= word2.size(); ++j){
  9. if(word1[i-1] == word2[j-1]){
  10. dp[i][j] = dp[i-1][j-1];
  11. }else{
  12. dp[i][j] = min(dp[i-1][j],min(dp[i][j-1],dp[i-1][j-1])) + 1;
  13. }
  14. }
  15. }
  16. return dp[word1.size()][word2.size()];
  17. }
  18. };

总结

题型:动态规划,编辑距离

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