leetcode72. Edit Distance
Posted LaplaceAkuir
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Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
给定任意两个字符串,要求只能用删除,替换,添加操作,将word1转变成word2,求最少操作次数。
声明dp[i][j],表示长度为i的字符串word1和长度为j的字符串word2匹配所需要的最小次数。
dp[i][j]有三种迭代可能:
1:dp[i][j] = dp[i-1][j] +1//1表示i多出了一个长度,需要删除。
2:dp[i][j] = dp[i][j-1]+1;//1表示i和j-1匹配的情况下,j多一个字符,需要添加这一步操作。
3:dp[i][j] = dp[i-1][j-1]+(word1[i-1]==word(j-1)?0:1);//表明在各自基础上添加一个字符,若相等则不操作,若不相等。则添加1步替换。
dp[i][j]=min(1,2,3);
1 class Solution { 2 public: 3 int minDistance(string word1, string word2) { 4 int m = word1.length(); 5 int n = word2.length(); 6 vector<vector<int> >dp(m+1,vector<int>(n+1)); 7 for(int i = 0; i <= m ;i++){ 8 for(int j = 0 ;j <= n ;j++){ 9 if(i == 0){ 10 dp[i][j] = j; 11 } 12 else if(j==0){ 13 dp[i][j] = i; 14 } 15 else{ 16 dp[i][j] = min(dp[i-1][j]+1,min(dp[i][j-1]+1,dp[i-1][j-1]+(word1[i-1]==word2[j-1]?0:1))); 17 } 18 } 19 } 20 return dp[m][n]; 21 } 22 };
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