LeetCode 583 两个字符串的删除操作

• 分析
  dp[i][j]表示word1的前i个字符和word2的前j个字符编程相同的字符串所需要的最小的操作数。状态转移方程见代码。
• 代码

class Solution {
    
    
public:
    int minDistance(string word1, string word2) {
    
    
        int len1 = word1.length();
        int len2 = word2.length();

        vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0));
        for(int i = 0; i <= len1; i++) dp[i][0] = i;
        for(int i = 0; i <= len2; i++) dp[0][i] = i;

        for(int i = 1; i <= len1; i++){
    
    
            for(int j = 1; j <= len2; 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[len1][len2];
    }
};