【代码随想录】day55

提示：文章写完后，目录可以自动生成，如何生成可参考右边的帮助文档

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();
    }
};