问题描述 链接到标题

72.编辑距离

解题思路 链接到标题

dp[i][j]的含义不再赘述:

  • if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
  • else,分为三种操作情况:
    • 替换末尾字符: dp[i][j] = dp[i - 1][j - 1] + 1;
    • 删除word1的第i个字符: dp[i][j] = dp[i - 1][j] + 1;
    • 删除word2的第j个字符，即相当于在第i个字符后插入word2[j - 1]: dp[i][j] = dp[i][j - 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 = 1; i <= word1.size(); i++) {
            dp[i][0] = i;
        }
        for (int j = 1; 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(min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]) + 1;
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }
};