Day 55 动态规划

392. 判断子序列

动态规划

就是用求最长公共子序列的方法，最后在返回时判断长度

class Solution {
public:bool isSubsequence(string s, string t) {int m = s.size() + 1, n = t.size() + 1;vector<vector<int>> dp(m, vector<int>(n, 0));for (int i = 1; i < m; i++){for (int j = 1; j < n; j++){if (s[i - 1] == t[j - 1]){dp[i][j] = dp[i - 1][j - 1] + 1;}else{// dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); // 只需要看左边的就行了，不用看上面的dp[i][j] = dp[i][j - 1]; // 等于字符串t前一位的情况}}}return dp[s.size()][t.size()] == s.size();}
};

双指针

双指针的方法似乎更加简单一点。

class Solution {
public:bool isSubsequence(string s, string t) {int sp = 0, tp = 0;while (sp < s.size() && tp < t.size()){if (s[sp] == t[tp]){sp++;}tp++;}return sp == s.size();}
};

115. 不同的子序列

class Solution {
public:int numDistinct(string s, string t) {int m = s.size(), n = t.size();vector<vector<uint64_t>> dp(m + 1, vector<uint64_t>(n + 1, 0));for (int i = 0; i <= m; i++){dp[i][0] = 1;}for (int i = 1; i <= m; i++){for (int j = 1; j <= n; j++){if (s[i - 1] == t[j - 1]){dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];}else{dp[i][j] = dp[i - 1][j];}}}return dp[m][n];}
};