顿搜
飞过闲红千叶，夕岸在哪
类目归类
[LeetCode 583] Delete Operation for Two Strings [Java] [Beats : 95.08%]
09/24
1. Description
Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.
Note:
The length of given words won't exceed 500.
Characters in given words can only be lower-case letters.
2. Runtime Distribution
3. Submission Details
4. Example
Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
5. Explanation
6. Code
public int minDistance(String word1, String word2) {
if (word1 == null || word2 == null) {
return 0;
}
int len1 = word1.length();
int len2 = word2.length();
int[][] dp = new int[len1 + 1][len2 + 1];
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] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return len1 + len2 - 2 * dp[len1][len2];
}7.Test
public class LeetCode0583 {
public int minDistance(String word1, String word2) {
if (word1 == null || word2 == null) {
return 0;
}
int len1 = word1.length();
int len2 = word2.length();
int[][] dp = new int[len1 + 1][len2 + 1];
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] + 1;
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return len1 + len2 - 2 * dp[len1][len2];
}
public static void main(String[] args) {
LeetCode0583 leetcode = new LeetCode0583();
System.out.println(leetcode.minDistance("sea", "eat"));
}
}
