2022比赛总结

牛客小白赛 / leetcode双周赛：

知识点总结：

1.__builtin_popcount(r),算数字里有几个1

class Solution {
public:
  vector<int> sortByBits(vector<int>& arr) {
    sort(arr.begin(), arr.end(), [](int l, int r) {
      int ln = __builtin_popcount(l);
      int rn = __builtin_popcount(r);
      if (ln != rn) return ln < rn;
      return l < r;
    });
   return arr;
  }
};

2.若直接计算数字中1的个数，可以用取余，除法来算

3.在类外直接写比较函数，加const &

4.第三题题解

后面的题不要抱有幻想，一定要用DP DFS/BFS之类的做，不要尝试暴力  

1358. Number of Substrings Containing All Three Characters

My Submissions Back to Contest

 

• User Accepted:958
• User Tried:1490
• Total Accepted:1003
• Total Submissions:2820
• Difficulty:Medium

Given a string s consisting only of characters a, b and c.

Return the number of substrings containing at least one occurrence of all these characters a, b and c.

Example 1:

Input: s = "abcabc"
Output: 10
Explanation: The substrings containing at least one occurrence of the characters a, b and c are "abc", "abca", "abcab", "abcabc", "bca", "bcab", "bcabc", "cab", "cabc" and "abc" (again). 

Example 2:

Input: s = "aaacb"
Output: 3
Explanation: The substrings containing at least one occurrence of the characters a, b and c are "aaacb", "aacb" and "acb".

Example 3:

Input: s = "abc"
Output: 1

Constraints:

• 3 <= s.length <= 5 x 10^4
• s only consists of a, b or c characters.

//sequence "abcabc" find substring contain 'a','b','c'
class Solution {
public:
    int numberOfSubstrings(string s) {
        int n = s.size();
        //two-dimension vector store range
        vector<vector<int> > dp(3, vector<int>(n + 1, -1));
        //store the nearest index of appearance a b c
        for (int i = n - 1; i >= 0; i--) {
            //no new appear record past
            for (int j = 0; j < 3; j++) dp[j][i] = dp[j][i + 1];
            //has new flush
            dp[s[i] - 'a'][i] = i;
        }
        long long ans = 0;
        for (int i = 0; i < n; i++) {
            bool ok = true;
            //if has -1 means lack of val
            for (int j = 0; j < 3; j++) {
                if (dp[j][i] == -1) ok = false;
            }
            //if ensure have three kinds calc the max distance to end
            if (ok) {
                int mx = -1;
                for (int j = 0; j < 3; j++) mx = max(mx, dp[j][i]);
                ans += n - mx;
            }
        }
        return ans;
    }
};