• 代码随想录 -- day56 -- 583. 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 

    1. class Solution {
    2. public:
    3. int minDistance(string word1, string word2) {
    4. vectorint>> 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] + 1, dp[i][j - 1] + 1);
    13. }
    14. }
    15. }
    16. return dp[word1.size()][word2.size()];
    17. }
    18. };

    72. 编辑距离 

    1. class Solution {
    2. public:
    3. int minDistance(string word1, string word2) {
    4. vectorint>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0));
    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. }
    12. else {
    13. dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
    14. }
    15. }
    16. }
    17. return dp[word1.size()][word2.size()];
    18. }
    19. };

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  • 原文地址:https://blog.csdn.net/Accelerated/article/details/133097871