LeetCode 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

class Solution {
public:
    int minDistance(string word1, string word2) {
        /*
        int m = word1.size(), n = word2.size();
        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
        for(int i = 0; i<=m; i++)
            dp[i][0] = i;
        for(int i = 1; i<=n; i++)
            dp[0][i] = i;
        for(int i = 1; i<=m; i++){
            for(int j = 1; j<=n; j++){
                int a = word1[i-1] == word2[j-1] ? 0 : 1;
                dp[i][j] = min(dp[i-1][j-1] + a, min(dp[i-1][j] + 1, dp[i][j-1] + 1));
            }
        }
        return dp[m][n];
        */
        int m = word1.size(), n = word2.size();
        vector<int> dp(n+1, 0);
        for(int i = 0; i<=n; i++)
            dp[i] = i;
        for(int i = 1; i<=m; i++){
            int pre = dp[0];
            dp[0] = i;
            for(int j = 1; j<=n; j++){
                int tmp = dp[j];
                int a = word1[i-1] == word2[j-1] ? 0 : 1;
                dp[j] = min(pre + a, min(dp[j] + 1, dp[j-1] + 1));
                pre = tmp;
            }
        }
        return dp[n];
    }
};

注：
1.注意dp[m][n]对dp[m-1][n],dp[m][n-1],dp[m-1][n-1]的依赖性。
2.滚动数组优化。