和查找子序列的操作类似,但是考虑的是删除操作
- class Solution {
- public:
- int minDistance(string word1, string word2) {
- vector
int>> dp(word1.size() + 1, vector<int>(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;
- 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()];
- }
- };
- class Solution {
- public:
- int minDistance(string word1, string word2) {
- vector
int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0)); - for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
- for (int j = 0; 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 - 1][j], dp[i][j - 1]}) + 1;
- }
- }
- }
- return dp[word1.size()][word2.size()];
- }
- };