杏坛网站建设深圳专业学校网站建设

思路

1. dp数组定义：0_i-1的字符串和0_j-1的字符串删除到相等时的最小步数
2. 递推公式：

   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;
   }

   如果相同时，代表不删除元素就行；不同时，需要选择删除i或者j中的其中一个元素，并且最小步数加一

3. dp数组初始化：

   vector< vector<int> > dp(word1.size() + 1, vector<int>(word2.size() + 1, INT_MAX));
   for(int i = 0; i <= word1.size(); i++){dp[i][0] = i;
   }
   for(int j = 0; j <= word2.size(); j++){dp[0][j] = j;
   }

4. 遍历顺序：顺序
5. 时间复杂度：      

代码

class Solution {
public:int minDistance(string word1, string word2) {vector< vector<int> > dp(word1.size() + 1, vector<int>(word2.size() + 1, INT_MAX));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()];}
};