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一、392判断子序列

双指针：

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

动态规划：

class Solution {
public:bool isSubsequence(string s, string t) {vector<int> dp(t.size() + 1, 0);vector<int> prev = dp;for (int i = 0; i < s.size(); i ++) {for (int j = 0; j < t.size(); j ++) {if (s[i] == t[j]) {dp[j+1] = prev[j] + 1;}else {dp[j+1] = max(prev[j+1], dp[j]);}}prev = dp;}return dp.back() == s.size();}
};

二、115不同的子序列

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

状态压缩（用int也行，回头再说吧，先写明白这种无脑复制数组版）：

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