**Leetcode 72. Edit Distance

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
time limit exceeded

class Solution {public:    int minDistance(string word1, string word2) {        size_t len1 = word1.size();        size_t len2 = word2.size();        if (len1 == 0) return len2;        if (len2 == 0) return len1;        int ret = max(len1, len2);        if (word1[0] == word2[0]) {            ret = min(ret, minDistance(string(word1.begin() + 1, word1.end()), string(word2.begin() + 1, word2.end())));        } else {            ret = min(ret, 1 + minDistance(string(word1.begin() + 1, word1.end()), string(word2.begin() + 1, word2.end()))); // replace            ret = min(ret, 1 + minDistance(string(word1.begin() + 1, word1.end()), word2)); // insert;            ret = min(ret, 1 + minDistance(word1, string(word2.begin() + 1, word2.end()))); // delete        }        return ret;    }};

这个题自己没有做出来， 没有想到动态规划，可见还是对动态规划算法的理解不够深入。
参考后

class Solution {public:    int minDistance(string word1, string word2) {        size_t r = word1.size();        size_t c = word2.size();        vector<vector<int> > dp(r + 1, vector<int>(c + 1, 0));        for (size_t i = 0; i <= r; ++i) {            dp[i][0] = i;        }        for (size_t i = 1; i <= c; ++i) {            dp[0][i] = i;        }        for (size_t i = 0; i != r; ++i) { // word1            for (size_t j = 0; j != c; ++j) { // word2                if (word1[i] == word2[j]) {                    dp[i + 1][j + 1] = dp[i][j];                } else {                    // insert dp[i + 1][j + 1] = dp[i + 1][j] + 1;                    // delete dp[i + 1][j + 1] = dp[i][j + 1] + 1;                    // replace dp[i + 1][j + 1] = dp[i][j] + 1;                    dp[i + 1][j + 1] = min(dp[i + 1][j] + 1, dp[i][j + 1] +  1);                    dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + 1);                }            }        }        return dp[r][c];    }};