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
题目大意是: 有两个 字符串, 找到把word1 转化成 word2 最少的操作步骤。
维护一个二维数组 dp[ i ][ j ] , dp[ i ][ j ] 表示 把 word1 [0 .. i - 1] 转换到 word1 [0 .. j - 1] 所需的最小步骤。
例子: word1: abcd word2: qwer
dp[ 4 ][ 0 ] 表示 把 abcd 转化成 “” 所需 步骤, 所以是 4.
所以
dp[ i ][ 0 ] = i
dp[
0 ][ j ] = j
如果 word1[ i - 1] 等于 word2[ j - 1] ,dp[ i ][ j ] = dp[ i - 1][ j - 1] ,则不需要做任何改动
If they are not equal, we need to consider three cases:
- Replace
word1[i - 1]
byword2[j - 1]
(dp[i][j] = dp[i - 1][j - 1] + 1 (for replacement)
); - Delete
word1[i - 1]
andword1[0..i - 2] = word2[0..j - 1]
(dp[i][j] = dp[i - 1][j] + 1 (for deletion)
); - Insert
word2[j - 1]
toword1[0..i - 1]
andword1[0..i - 1] + word2[j - 1] = word2[0..j - 1]
(dp[i][j] = dp[i][j - 1] + 1 (for insertion)
).
替换:把word1[ i - 1] 替换成 word2[ j - 1] 则需要 把 在dp[ i - 1][ j - 1]的基础上 加 1, 所以是 dp[ i ][ j ] = dp[ i - 1][ j - 1] + 1
删除:把word1[ i - 1] 删除, 使得 word1[ i - 2] 和 word2[ j - 1] 一样, 需要把替换成 word2[ j - 1] , dp[ i ][ j ] = dp[ i - 1][ j ] + 1
加入:把word2[ j - 1] 加入到 word1中, dp[ i ][ j ] = dp[ i ][ j - 1] + 1
public int minDistance(String word1, String word2) {
int len1 = word1.length(), len2 = word2.length();
int[][] dp = new int[len1 + 1][len2 + 1];
for (int i = 1; i <= len1; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= len2; j++) {
dp[0][j] = j;
}
for (int i = 1; i <= len1; i++) {
for (int j = 1; j <= len2; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1];
else dp[i][j] = Math.min(dp[i][j - 1], Math.min(dp[i - 1][j - 1], dp[i - 1][j])) + 1;
}
}
return dp[len1][len2];
}