题目1:115. 不同的子序列 - 力扣(LeetCode)
这里的初始化第一列为1,表示t是空字符串 在s的子序列中有1种情况
递推公式也不同
class Solution {
public:
int numDistinct(string s, string t) {
vector<vector<int>> dp(s.size() + 1, vector<int>(t.size() + 1));
for(int i = 0;i <= s.size();i++) {
dp[i][0] = 1;
}
for(int i = 1;i <= s.size();i++) {
for(int j = 1;j <= t.size();j++) {
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];
}
}
return dp[s.size()][t.size()];
}
};
题目2:583. 两个字符串的删除操作 - 力扣(LeetCode)
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
for(int i = 1;i <= word1.size();i++) {
dp[i][0] = i;
}
for(int j = 1;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][j - 1] + 1);
}
}
}
return dp[word1.size()][word2.size()];
}
};
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
for(int i = 1;i <= word1.size();i++) dp[i][0] = i;
for(int j = 1;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, min(dp[i][j - 1] + 1, dp[i - 1][j - 1] + 1));
}
}
}
return dp[word1.size()][word2.size()];
}
};