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
这道题目是关于有两个sequence 的dp 题目(最长公共自序列,也为two sequences),这类题目往往需要用一个二维数组,例如 f[ i ][ j ].
i 表示sequence 1 里的某个位置,j stands for sequence2 里的某个位置,然后进行匹配操作。
对于这道题目,重点是理解 function 为什么这么写(网上很多讲解):
num[i][j] = Math.min(Math.min(num[i - 1][j - 1], num[i][j - 1]),num[i - 1][j]) + 1;
public class Solution {
public int minDistance(String word1, String word2) {
if (word1 == null && word2 == null) {
return -1;
}
int l1 = word1.length();
int l2 = word2.length();
if (l1 == 0) {
return l2;
}
if (l2 == 0) {
return l1;
}
int[][] num = new int[l1+1][l2+1];
for (int i = 0; i <= l1; i++) {
num[i][0] = i;
}
for (int j = 1; j <= l2; j++) {
num[0][j] = j;
}
for (int i = 1; i <= l1; i++) {
for (int j = 1; j <= l2; j++) {
if (word1.charAt(i-1) == word2.charAt(j-1)) {
num[i][j] = num[i - 1][j - 1];
} else {
num[i][j] = Math.min(Math.min(num[i - 1][j - 1], num[i][j - 1]),num[i - 1][j]) + 1;
}
}
}
return num[l1 ][l2 ];
}
}
注意,当
word1.charAt(i - 1) == word2.charAt(j - 1)
时,当前字符不需要编辑, 直接返回num[i][j]即可
同时,对比第一个代码可以看到,当 word1.length() == 0 或者word2.length()== 0 是不需要拿出来单独考虑的。
public class Solution {
public int minDistance(String word1, String word2) {
if (word1 == null || word2 == null) {
return -1;
}
int[][] num = new int[word1.length() + 1][word2.length() + 1];
num[0][0] = 0;
for (int i = 1; i <= word2.length(); i++) {
num[0][i] = i;
}
for (int i = 1; i <= word1.length(); i++) {
num[i][0] = i;
}
for (int i = 1; i <= word1.length(); i++) {
for (int j = 1; j <= word2.length(); j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
num[i][j] = num[i - 1][j - 1];
} else {
int delete_insert = Math.min(num[i - 1][j], num[i][j - 1]) ;
num[i][j] = Math.min(delete_insert, num[i - 1][j - 1]) + 1;
}
}
}
return num[word1.length()][word2.length()];
}
}
本文详细介绍了如何使用动态规划解决字符串转换问题,通过构建二维数组并运用数学公式进行状态转移,最终计算出将一个字符串转换为另一个字符串所需的最少步骤。文章包括代码实现和关键逻辑解释。
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