Edit Distance(动态规划)
Edit Distance
更多技术博客 http://vilins.top/
题目
Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:
Insert a character
Delete a character
Replace a character
Example 1:
Input: word1 = “horse”, word2 = “ros”
Output: 3
Explanation:
horse -> rorse (replace ‘h’ with ‘r’)
rorse -> rose (remove ‘r’)
rose -> ros (remove ‘e’)
Example 2:
Input: word1 = “intention”, word2 = “execution”
Output: 5
Explanation:
intention -> inention (remove ‘t’)
inention -> enention (replace ‘i’ with ‘e’)
enention -> exention (replace ‘n’ with ‘x’)
exention -> exection (replace ‘n’ with ‘c’)
exection -> execution (insert ‘u’)
分析
用动态规划的思想去解决这个问题，那么意味着我们需要寻找这两个字符串的子问题的解，那么这个子问题怎么解决呢？
我们采用一个二维数组来表示count[i][j]表示字符串长为i和字符串长为j的字符的编辑距离，对于子问题，我们可以采用将i和j逐步缩小的策略，那么最终问题可以看作解决count[i][j]，而每个字符串与空串的编辑距离是明显的，及count[0][j]和count[i][0]是明显知道的。
那么对于子问题可以有一下三种情况
最后可以知道状态转移方程为：
其中diff(i,j)是指当串1和串2的下标i和下标j对应的字符相等的时候为0，否则需要一步编辑距离，则为1.
源码
class Solution { public: int minDistance(string word1, string word2) { int size1 = word1.size(); int size2 = word2.size(); vector<vector<int>> count(size1+1, vector<int>(size2+1, 0)); for(int i = 0; i <= size1; i++) { count[i][0] = i; } for(int i = 0; i <= size2; i++) { count[0][i] = i; } for(int i = 1; i <= size1; i++) { for(int j = 1; j <= size2; j++) { int tag = 0; if(word1[i-1] != word2[j-1]) { tag = 1; } count[i][j] = min(count[i-1][j]+1, count[i][j-1]+1); count[i][j] = min(count[i][j], tag+count[i-1][j-1]); } } return count[size1][size2]; } };觉得博主写的好