刷题顺序以及题解参考卡哥的代码随想录
题目描述
英文版描述
Given two strings word1 and word2, return the minimum number of steps required to makeword1andword2the same.
In one step, you can delete exactly one character in either string.
Example 1:
Input: word1 = "sea", word2 = "eat" Output: 2 Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
Example 2:
Input: word1 = "leetcode", word2 = "etco" Output: 4
Constraints:
- 1 <= word1.length, word2.length <= 500
- word1 and word2 consist of only lowercase English letters.
英文版地址
中文版描述
给定两个单词 word1 和 word2 ，返回使得 word1 和 word2相同所需的最小步数。
每步 可以删除任意一个字符串中的一个字符。
示例 1：
输入: word1 = "sea", word2 = "eat" 输出: 2 解释: 第一步将 "sea" 变为 "ea" ，第二步将 "eat "变为 "ea"
示例 2:
输入： word1 = "leetcode", word2 = "etco" 输出： 4
提示：
- 1 <= word1.length, word2.length <= 500
- word1 和 word2 只包含小写英文字母
中文版地址
解题方法
class Solution {
public int minDistance(String word1, String word2) {
int[][] dp = new int[word1.length()][word2.length()];
if (word1.charAt(0) != word2.charAt(0)) {
dp[0][0] = 2;
}
for (int i = 1; i < word1.length(); i++) {
if (word1.charAt(i) == word2.charAt(0) && dp[i - 1][0] != i-1) {
dp[i][0] = dp[i - 1][0] - 1;
} else {
dp[i][0] = dp[i - 1][0] + 1;
}
}
for (int j = 1; j < word2.length(); j++) {
if (word1.charAt(0) == word2.charAt(j)) {
dp[0][j] = j;
} else {
dp[0][j] = dp[0][j - 1] + 1;
}
}
for (int i = 1; i < word1.length(); i++) {
for (int j = 1; j < word2.length(); j++) {
if (word1.charAt(i) == word2.charAt(j)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i][j - 1], dp[i - 1][j]) + 1;
}
}
}
return dp[word1.length() - 1][word2.length() - 1];
}
}
复杂度分析
- 时间复杂度：O(n*m)，其中 n 是word1的长度，m 是word2的长度
- 空间复杂度：O(n*m)
