[Leetcode] Edit Distance

题目：

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

思路：DP. 状态转移方程需要分情况讨论，对于word1[0...i] 和 word2[0...j]，需要比较两个字符串的最后一位，若相等，那么一种可能的变换方式为先进行word1[0...i-1] 和 word2[0...j-1] 的转换，然后加上结尾的字符；在这种情况下，distance与dp[i-1][j-1]相同。如果最后一位不等，那么对应的变换方式为先进行word1[0...i-1] 和 word2[0...j-1] 的转换，再进行最后一位的变换；在这种情况下，distance为dp[i-1][j-1] + 1. 另外还有两种trivial的变换方式，即从word1[0...i-1] 变换到 word2[0...j]，之后在word1结尾加一个字符；和从word1[0...i] 变换到 word2[0...j-1]，之后在word2结尾加一个字符。最终的结果，即为三种变换方式中最小的那个。

class Solution {public:    int min(int a, int b, int c) {        int temp = a < b ? a : b;        return temp < c ? temp : c;    }        int minDistance(string word1, string word2) {        int len1 = (int)word1.size();        int len2 = (int)word2.size();        vector<vector<int>> dist(len1 + 1, vector<int>(len2 + 1, 0));        //dist[i][j] : steps needed to convert from word1[1~i] to word2[1~j]        for (int i = 0; i <= len1; ++i) {            dist[i][0] = i;        }        for (int j = 0; j <= len2; ++j) {            dist[0][j] = j;        }        for (int i = 1; i <= len1; ++i) {            for (int j = 1; j <= len2; ++j) {                if (word1[i-1] == word2[j-1]) {                    dist[i][j] = min(dist[i-1][j-1], dist[i-1][j] + 1, dist[i][j-1] + 1);                } else {                    dist[i][j] = min(dist[i-1][j-1] + 1, dist[i-1][j] + 1, dist[i][j-1] + 1);                }            }        }        return dist[len1][len2];    }};

总结：复杂度为O(mn).