583. 两个字符串的删除操作 - 力扣(LeetCode)
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
class Solution {
public:
int minDistance(string word1, string word2) {
vector> dp(word1.size()+1, vector(word2.size()+1, 0));
for (int i=1; i<=word1.size(); i++){
for (int j=1; j<=word2.size(); j++){
if (word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
}
}
return word1.size()+word2.size()-dp[word1.size()][word2.size()]*2;
}
};
解法代码:代码随想录 (programmercarl.com)
72. 编辑距离 - 力扣(LeetCode)
给你两个单词 word1 和 word2, 请返回将 word1 转换成 word2 所使用的最少操作数 。
你可以对一个单词进行如下三种操作:
插入一个字符
删除一个字符
替换一个字符
示例 1:
输入:word1 = "horse", word2 = "ros"
输出:3
解释:
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')
示例 2:
输入:word1 = "intention", word2 = "execution"
输出:5
解释:
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')
提示:
0 <= word1.length, word2.length <= 500
word1 和 word2 由小写英文字母组成
class Solution {
public:
int minDistance(string word1, string word2) {
vector> dp(word1.size() + 1, vector(word2.size() + 1, 0));
for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
for (int i = 1; i <= word1.size(); i++) {
for (int j = 1; j <= word2.size(); j++) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
}
else {
dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
}
}
}
return dp[word1.size()][word2.size()];
}
};
解法代码:代码随想录 (programmercarl.com)
代码随想录 (programmercarl.com)