Add files via upload

Author: assassint2017

Longest Substring with At Least K Repeating Characters/Longest_Substring_with_At_Least_K_Repeating_Characters.cpp

@@ -0,0 +1,73 @@
+// 使用递归的思想，附上大神的解题思路，是在是太NB了
+// https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/discuss/87768/4-lines-Python
+
+// Runtime: 0 ms, faster than 100.00% of C++ online submissions for Longest Substring with At Least K Repeating Characters.
+// Memory Usage: 9.5 MB, less than 27.27% of C++ online submissions for Longest Substring with At Least K Repeating Characters.
+
+class Solution
+{
+public:
+	int longestSubstring(string s, int k)
+	{
+        // 递归停止条件
+		if (k <= 0 || k > s.length())
+			return 0;
+
+		vector<int> hashmap(128, 0);
+		for (char chr : s)
+			++hashmap[chr];
+
+		for (char chr = 'a'; chr <= 'z'; ++chr)
+		{
+			if (hashmap[chr] < k && hashmap[chr] > 0)
+			{
+				int res = 0;
+				for (int leftPtr = 0, rightPtr = 0; leftPtr < s.length(); leftPtr = rightPtr + 1, rightPtr = leftPtr)
+				{
+					while (rightPtr < s.length() && s[rightPtr] != chr)
+						++rightPtr;
+
+					res = max(res, longestSubstring(s.substr(leftPtr, rightPtr - leftPtr), k));
+				}
+
+				return res;
+			}
+		}
+
+		return s.length();
+	}
+};
+
+// 下面是自己写的真n2的时间复杂度的，果然是非常的慢
+// Runtime: 636 ms, faster than 5.05% of C++ online submissions for Longest Substring with At Least K Repeating Characters.
+// Memory Usage: 10.9 MB, less than 27.27% of C++ online submissions for Longest Substring with At Least K Repeating Characters.
+
+class Solution 
+{
+public:
+    int longestSubstring(string s, int target) 
+    {
+        int res = 0;
+        for (int i = 0; i < s.length(); ++i)    
+        {
+            vector<int> hashmap(26, 0);
+            for (int j = i; j < s.length(); ++j)
+            {
+                ++hashmap[s[j] - 'a'];
+                
+                bool ok = true;
+                for (int k = 0; k < 26; ++k)
+                {
+                    if (hashmap[k] > 0 && hashmap[k] < target)
+                    {
+                        ok = false;
+                        break;
+                    }
+                }
+                
+                if (ok) res = max(res, j - i + 1);
+            }
+        }
+        return res;
+    }
+};