//time:o(mn)

//space:o(mn)

class Solution {

public int minDistance(String word1, String word2) {

int n1 = word1.length();

int n2 = word2.length();

int[][] dp = new int[n1 + 1][n2 + 1];

// dp[0][0...n2]的初始值

for (int j = 1; j <= n2; j++)

dp[0][j] = dp[0][j - 1] + 1;

// dp[0...n1][0] 的初始值

for (int i = 1; i <= n1; i++)

dp[i][0] = dp[i - 1][0] + 1;

// 通过公式推出 dp[n1][n2]

for (int i = 1; i <= n1; i++) {

for (int j = 1; j <= n2; j++) {

// 如果 word1[i] 与 word2[j] 相等。第 i 个字符对应下标是 i-1

if (word1.charAt(i - 1) == word2.charAt(j - 1)){

dp[i][j] = dp[i - 1][j - 1];

}else {

dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]) + 1;

}

}

}

return dp[n1][n2];

}

}