leetcode 395. Longest Substring with At Least K Repeating Characters 最长K个数量的字符 + DFS深度优先搜索 + 移动窗口

以你之姓@ 2022-06-04 06:25 116阅读 0赞

Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times.

Example 1:

Input:  
s = “aaabb”, k = 3

Output:  
3

The longest substring is “aaa”, as ‘a’ is repeated 3 times.  
Example 2:

Input:  
s = “ababbc”, k = 2

Output:  
5

The longest substring is “ababb”, as ‘a’ is repeated 2 times and ‘b’ is repeated 3 times.

本题题意是说寻找一个字符子串，要求每一个字符至少出现k次，求这个子串的最大长度，这个是网上看的做法，基本思路如下：首先先遍历统计字符的数量，然后寻找第一个出现不超过k次的字符，从这个字符分成左右两个子串，递归判断，直至找到最大值

可能是test case 更新了，所以递归做法会超时，于是后来又看了一个做法，

由于字母只有26个，而整型mask有32位，足够用了，每一位代表一个字母，如果为1，表示该字母不够k次，如果为0就表示已经出现了k次，这种思路真是太聪明了。我们遍历字符串，对于每一个字符，我们都将其视为起点，然后遍历到末尾，我们增加哈希表中字母的出现次数，如果其小于k，我们将mask的对应位改为1，如果大于等于k，将mask对应位改为0。然后看mask是否为0，是的话就更新res结果，然后把当前满足要求的子字符串的起始位置j保存到max\_idx中，等内层循环结束后，将外层循环变量i赋值为max\_idx+1，继续循环直至结束，

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    class Solution
    {
    public:
        int longestSubstring(string s, int k)
        {
            if (k <= 0)
                return s.length();
            if (s.length() <= 0 || k > s.length())
                return 0;
    
            int i = 0, maxLen = 0;
            while (i + k <= s.length())
            {
                int mask = 0, index = i;;
                vector<int> count(26, 0);
                for (int j = i; j < s.length(); j++)
                {
                    count[s[j] - 'a']++;
                    if (count[s[j] - 'a'] < k)
                        mask |= (1 << (s[j] - 'a'));
                    else
                        mask &= (~(1 << (s[j] - 'a')));
                    if (mask == 0)
                    {
                        maxLen = max(maxLen, j - i + 1);
                        index = j;
                    }
                }
                i = index + 1;
            }
            return maxLen;
        }
    
        int longestSubstringByDFS(string s, int k)
        {
            if (k <= 0)
                return s.length();
            else if (s.length() <= 0 || k > s.length())
                return 0;
            else
            {
                map<char, int> mmp;
                for (char a : s)
                {
                    if (mmp.find(a) == mmp.end())
                        mmp[a] = 1;
                    else
                        mmp[a] += 1;
                }
    
                int index = 0;
                for (index = 0; index<s.length(); index++)
                {
                    if (mmp[s[index]] < k)
                        break;
                }
                if (index == s.length())
                    return s.length();
                else
                {
                    string left = s.substr(0, index);
                    string right = s.substr(index + 1);
                    int a = longestSubstring(left, k);
                    int b = longestSubstring(right, k);
                    return max(a, b);
                }
            }
        }
    };