刷题的第四十一天,希望自己能够不断坚持下去,迎来蜕变。😀😀😀
刷题语言:C++
Day41 任务
● 583. 两个字符串的删除操作
● 72. 编辑距离
● 编辑距离总结篇
1 两个字符串的删除操作
583. 两个字符串的删除操作
思路:
动态规划
(1)确定dp数组(dp table)以及下标的含义
dp[i][j]:以i-1为结尾的字符串word1,和以j-1为结尾的字符串word2,想要达到相等,所需要删除元素的最少次数。
(2)确定递推公式
- 当word1[i - 1] 与 word2[j - 1]相同的时候
d p [ i ] [ j ] = d p [ i − 1 ] [ j − 1 ] ; dp[i][j] = dp[i - 1][j - 1]; dp[i][j]=dp[i−1][j−1]; - 当word1[i - 1] 与 word2[j - 1]不相同的时候
① 删word1[i - 1],最少操作次数为dp[i - 1][j] + 1
② 删word2[j - 1],最少操作次数为dp[i][j - 1] + 1
③ 同时删word1[i - 1]和word2[j - 1],操作的最少次数为dp[i - 1][j - 1] + 2
递推公式: d p [ i ] [ j ] = m i n ( d p [ i − 1 ] [ j − 1 ] + 2 , d p [ i − 1 ] [ j ] + 1 , d p [ i ] [ j − 1 ] + 1 ) ; dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1}); dp[i][j]=min(dp[i−1][j−1]+2,dp[i−1][j]+1,dp[i][j−1]+1);
(3)dp数组如何初始化
vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
(4)确定遍历顺序
从上到下,从左到右
(5)举例推导dp数组
C++:
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
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] + 2, min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));
}
}
}
return dp[word1.size()][word2.size()];
}
};
时间复杂度:
O
(
n
∗
m
)
O(n * m)
O(n∗m)
空间复杂度:
O
(
n
∗
m
)
O(n * m)
O(n∗m)
2 编辑距离
72. 编辑距离
思路:
动态规划
(1)确定dp数组(dp table)以及下标的含义
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]
(2)确定递推公式
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])
dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
(3)dp数组如何初始化
for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
(4)确定遍历顺序:从左到右从上到下去遍历
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;
}
}
}
(5)举例推导dp数组
C++:
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][j - 1], dp[i - 1][j], dp[i - 1][j - 1]}) + 1;
}
}
}
return dp[word1.size()][word2.size()];
}
};
时间复杂度:
O
(
n
∗
m
)
O(n * m)
O(n∗m)
空间复杂度:
O
(
n
∗
m
)
O(n * m)
O(n∗m)
3 动态规划之编辑距离总结
判断子序列
if (s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
else dp[i][j] = dp[i][j - 1];
不同的子序列
if (s[i - 1] == t[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
} else {
dp[i][j] = dp[i - 1][j];
}
两个字符串的删除操作
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] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
}
编辑距离
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;
}
鼓励坚持四十二天的自己😀😀😀