LeetCode 72. Edit Distance(java)

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：动态规划解

DP: dp[i][j]代表word1的前i个和word2的前j个相同的最少操作步数。
case1 : (word1[i] == word2[j]) dp[i][j] = dp[i-1][j-1];
case2 : (word1[i] == word2[j]) insert: dp[i][j] = dp[i][j-1];
delete: dp[i][j] = dp[i-1][j];
replace: dp[i][j] = dp[i-1][j-1];
base case: dp[i][0] = dp[0][i] = i;

代码：

public int minDistance(String word1, String word2) {
if (word1.length() == 0) return word2.length();
if (word2.length() == 0) return word1.length();
int m = word1.length(), n = word2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i < m; i++) dp[i][0] = i;
for (int i = 0; i < n; i++) dp[0][i] = i;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1.charAt(i) == word2.charAt(j)) {
dp[i][j] = dp[i - 1][j -1 ];
} else {
int insert = dp[i][j - 1];
int delete = dp[i - 1][j];
int replace = dp[i -1][j - 1];
dp[i][j] = Math.min(insert, Math.min(delete, replace));
dp[i][j]++;
}
}
}
return dp[m]
;
}

//更加优化的dp方法，空间复杂度上的优化：通过一个数组和两个指针来代替m*n的二维数组空间，pre代表当前更新数的左上角的数，temp是事先预存下一个数的左上角的数，因此每次先更新为当前数的上面的数。
public int minDistance(String word1, String word2) {
if (word1.length() == 0) return word2.length();
if (word2.length() == 0) return word1.length();
int m = word1.length(), n = word2.length();
int[] dp = new int[n + 1];
for (int i = 0; i <= n; i++) dp[i] = i;
for (int i = 1; i <= m; i++) {
int temp = 0, pre = i - 1;
dp[0] = i;
for (int j = 1; j <= n; j++) {
temp = dp[j];
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[j] = pre;
} else {
dp[j] = Math.min(dp[j], Math.min(dp[j-1], pre));
dp[j]++;
}
pre = temp;
}
}
return dp
;
}