标签:++i lse solution https 初始 string 官方 删除 单词
给定两个单词?word1 和?word2,计算出将?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‘)
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/edit-distance
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
dp[i][j] = dp[i - 1][j - 1] ,word[i-1]=word2[j-1]
dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1] , dp[i - 1][j]), dp[i][j - 1]) + 1,word[i-1]!=word2[j-1]
for (int j = 0; j <= word2.length(); ++j) {
dp[0][j] = j;
}
for (int i = 0; i <= word1.length(); ++i) {
dp[i][0] = i;
}
class Solution {
public int minDistance(String word1, String word2) {
int[][] dp = new int[word1.length() + 1][word2.length() + 1];
for (int j = 0; j <= word2.length(); ++j) {
dp[0][j] = j;
}
for (int i = 0; i <= word1.length(); ++i) {
dp[i][0] = i;
}
for (int i = 1; i <= word1.length(); ++i) {
for (int j = 1; j <= word2.length(); ++j) {
if (word2.charAt(j - 1) == word1.charAt(i - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i - 1][j]), dp[i][j - 1]) + 1;
}
}
}
return dp[word1.length()][word2.length()];
}
}
标签:++i lse solution https 初始 string 官方 删除 单词
原文地址:https://www.cnblogs.com/coding-gaga/p/12267438.html