第72題，題目如下:
Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.
You have the following three operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
Example 1:
Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution" Output: 5 Explanation: intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')
Constraints:
0 <= word1.length, word2.length <= 500word1andword2consist of lowercase English letters.
採用動態規畫(dp)的方式，因為第一串文字不修改字元則計為0次，修改則最多為m次；第二串文字不修改字元時計為0次，修改則最多為n次。
dp選擇二維的陣列形式來進行，dp[m][n]與dp[m-1][n-1]的值相同時，不需修改，即dp[m][n] = dp[m-1][n-1]。其他時候，如果dp走更新，則dp[m][n] = dp[m-1][n-1]+1；走刪除，則dp[m][n] = dp[m-1][n]+1；如果走新增，則dp[m][n] = dp[m][n-1]+1；將此3種方式取最小值Math.min()來進行，得最佳解。
用這種方式來運作，最後回傳動態規畫結果dp[m][n]，即為題目所求的修改文字字串的最少變更次數值。
“`
public int minDistance(String word1, String word2) {
final int m = word1.length();
final int n = word2.length();
// dp[i][j] := the minimum number of operations to convert word1[0..i) to
// word2[0..j)
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i <= m; ++i) { dp[i][0] = i; }
for (int j = 1; j <= n; ++j) { dp[0][j] = j; }
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (word1.charAt(i – 1) == word2.charAt(j – 1)) // word char the same
dp[i][j] = dp[i – 1][j – 1];
else
dp[i][j] = Math.min(dp[i – 1][j – 1],
Math.min(dp[i – 1][j], dp[i][j – 1])) + 1;
}
}
return dp[m][n];
“`
This is great let me know more