1 class Solution {
 2 public:
 3     /**
 4      * @param word1 & word2: Two string.
 5      * @return: The minimum number of steps.
 6      */
 7     int minDistance(string word1, string word2) {
 8         // write your code here
 9         int m = word1.length(), n = word2.length();
10         vector<vector<int> > dp(m + 1, vector<int>(n + 1));
11         for (int i = 0; i <= m; ++i) {
12             for (int j = 0; j <= n; ++j) {
13                 if ( i== 0 ) {
14                     dp[i][j] = j;
15                 } else if (j == 0) {
16                     dp[i][j] = i;
17                 } else {
18                     dp[i][j] = min(dp[i - 1][j - 1] + ((word1[i - 1] == word2[j - 1])?0:1),
19                                    min(dp[i][j - 1] + 1, dp[i - 1][j] + 1)
20                                 );
21                 }
22             }
23         }
24         return dp[m][n];
25     }
26 };

 1 class Solution {
 2 public:
 3     /**
 4      * @param word1 & word2: Two string.
 5      * @return: The minimum number of steps.
 6      */
 7     int minDistance(string word1, string word2) {
 8         // write your code here
 9         int m = word1.length(), n = word2.length();
10         // vector<vector<int> > dp(m + 1, vector<int>(n + 1));
11         vector<int> dp(n + 1);
12         for (int i = 0; i <= m; ++i) {
13             int last;
14             for (int j = 0; j <= n; ++j) {
15                 if ( i== 0 ) {
16                     dp[j] = j;
17                 } else if (j == 0) {
18                     last = dp[j];
19                     dp[j] = i;
20                 } else {
21                     int tmp = dp[j];
22                     dp[j] = min(last + ((word1[i - 1] == word2[j - 1])?0:1),
23                                    min(dp[j - 1] + 1, dp[j] + 1)
24                                 );
25                     last = tmp;
26                 }
27             }
28         }
29         return dp[n];
30     }
31 };