导读 ^ _ ^
编辑距离是线性DP中比较困难的题目。
关键在于如何看待删，增，换。
题目
leetcode 72
代码与思路
确定状态
- dp[i][j] 表示以下标i-1为结尾的字符串word1，和以下标j-1为结尾的字符串word2，最近编辑距离为dp[i][j]。
if (word1[i - 1] == word2[j - 1])
不操作：dp[i][j] = dp[i - 1][j - 1];
if (word1[i - 1] != word2[j - 1])
增：增一个元素和删除一个元素一样
删：删前 dp[i][j] = dp[i - 1][j] + 1; 删后dp[i][j] = dp[i][j - 1] + 1;
换：前面的匹配好的再加当前一个替换数 dp[i][j] = dp[i - 1][j - 1] + 1;
综上：取最小 dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
变成空字符的操作数
for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
遍历答案
- 从左上推导到右下
//编辑距离
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> dp(word1.size() + 1, vector<int>(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()];
}
};