问题描述 链接到标题

583.两个字符串的删除操作

解题思路 链接到标题

dp[i][j]表示对word1的前i个字符，word2的前j个字符，使得它们相同的最小步数:

  • if (word1[i - 1] == word2[j - 1]), dp[i][j] = dp[i - 1][j - 1];
  • else, dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;

代码 链接到标题 

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 = 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], dp[i][j - 1]) + 1;
            }
        }
        return dp[word1.size()][word2.size()];
    }
};