Add solution 0395

leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/395. Longest Substring with At Least K Repeating Characters.go

@@ -0,0 +1,70 @@
+package leetcode
+
+import "strings"
+
+// 解法一 滑动窗口
+func longestSubstring(s string, k int) int {
+	res := 0
+	for t := 1; t <= 26; t++ {
+		freq, total, lessK, left, right := [26]int{}, 0, 0, 0, -1
+		for left < len(s) {
+			if right+1 < len(s) && total <= t {
+				if freq[s[right+1]-'a'] == 0 {
+					total++
+					lessK++
+				}
+				freq[s[right+1]-'a']++
+				if freq[s[right+1]-'a'] == k {
+					lessK--
+				}
+				right++
+			} else {
+				if freq[s[left]-'a'] == k {
+					lessK++
+				}
+				freq[s[left]-'a']--
+				if freq[s[left]-'a'] == 0 {
+					total--
+					lessK--
+				}
+				left++
+			}
+			if lessK == 0 {
+				res = max(res, right-left+1)
+			}
+
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 递归分治
+func longestSubstring1(s string, k int) int {
+	if s == "" {
+		return 0
+	}
+	freq, split, res := [26]int{}, byte(0), 0
+	for _, ch := range s {
+		freq[ch-'a']++
+	}
+	for i, c := range freq[:] {
+		if 0 < c && c < k {
+			split = 'a' + byte(i)
+			break
+		}
+	}
+	if split == 0 {
+		return len(s)
+	}
+	for _, subStr := range strings.Split(s, string(split)) {
+		res = max(res, longestSubstring1(subStr, k))
+	}
+	return res
+}

leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/395. Longest Substring with At Least K Repeating Characters_test.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question395 struct {
+	para395
+	ans395
+}
+
+// para 是参数
+// one 代表第一个参数
+type para395 struct {
+	s string
+	k int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans395 struct {
+	one int
+}
+
+func Test_Problem395(t *testing.T) {
+
+	qs := []question395{
+
+		{
+			para395{"aaabb", 3},
+			ans395{3},
+		},
+
+		{
+			para395{"ababbc", 2},
+			ans395{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 395------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans395, q.para395
+		fmt.Printf("【input】:%v       【output】:%v\n", p, longestSubstring(p.s, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0395.Longest-Substring-with-At-Least-K-Repeating-Characters/README.md

@@ -0,0 +1,112 @@
+# [395. Longest Substring with At Least K Repeating Characters](https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/)
+
+
+## 题目
+
+Given a string `s` and an integer `k`, return *the length of the longest substring of* `s` *such that the frequency of each character in this substring is greater than or equal to* `k`.
+
+**Example 1:**
+
+```
+Input: s = "aaabb", k = 3
+Output: 3
+Explanation: The longest substring is "aaa", as 'a' is repeated 3 times.
+```
+
+**Example 2:**
+
+```
+Input: s = "ababbc", k = 2
+Output: 5
+Explanation: The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 10^4`
+- `s` consists of only lowercase English letters.
+- `1 <= k <= 10^5`
+
+## 题目大意
+
+给你一个字符串 s 和一个整数 k ，请你找出 s 中的最长子串， 要求该子串中的每一字符出现次数都不少于 k 。返回这一子串的长度。
+
+## 解题思路
+
+- 最容易想到的思路是递归。如果某个字符出现次数大于 0 小于 k，那么包含这个字符的子串都不满足要求。所以按照这个字符来切分整个字符串，满足题意的最长子串一定不包含切分的字符。切分完取出最长子串即可。时间复杂度 O(26*n)，空间复杂度 O(26^2)
+- 此题另外一个思路是滑动窗口。有一个需要解决的问题是右窗口移动的条件。此题要求最长字符串，那么这个最终的字符串内包含的字符种类最多是 26 种。字符种类就是右窗口移动的条件。依次枚举字符种类，如果当前窗口内的字符种类小于当前枚举的字符种类，那么窗口右移，否则左移。窗口移动中需要动态维护 freq 频次数组。可以每次都循环一遍这个数组，计算出出现次数大于 k 的字符。虽然这种做法只最多循环 26 次，但是还是不高效。更高效的做法是维护 1 个值，一个用来记录当前出现次数小于 k 次的字符种类数 `less`。如果 freq 为 0 ，说明小于 k 次的字符种类数要发生变化，如果是右窗口移动，那么 `less++`，如果是左窗口移动，那么`less--`。同理，如果 freq 为 k ，说明小于 k 次的字符种类数要发生变化，如果是右窗口移动，那么 `less--`，如果是左窗口移动，那么`less++`。在枚举 26 个字符种类中，动态维护记录出最长字符串。枚举完成，最长字符串长度也就求出来了。时间复杂度 O(26*n)，空间复杂度 O(26)
+
+## 代码
+
+```go
+package leetcode
+
+import "strings"
+
+// 解法一 滑动窗口
+func longestSubstring(s string, k int) int {
+	res := 0
+	for t := 1; t <= 26; t++ {
+		freq, total, lessK, left, right := [26]int{}, 0, 0, 0, -1
+		for left < len(s) {
+			if right+1 < len(s) && total <= t {
+				if freq[s[right+1]-'a'] == 0 {
+					total++
+					lessK++
+				}
+				freq[s[right+1]-'a']++
+				if freq[s[right+1]-'a'] == k {
+					lessK--
+				}
+				right++
+			} else {
+				if freq[s[left]-'a'] == k {
+					lessK++
+				}
+				freq[s[left]-'a']--
+				if freq[s[left]-'a'] == 0 {
+					total--
+					lessK--
+				}
+				left++
+			}
+			if lessK == 0 {
+				res = max(res, right-left+1)
+			}
+
+		}
+	}
+	return res
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+// 解法二 递归分治
+func longestSubstring1(s string, k int) int {
+	if s == "" {
+		return 0
+	}
+	freq, split, res := [26]int{}, byte(0), 0
+	for _, ch := range s {
+		freq[ch-'a']++
+	}
+	for i, c := range freq[:] {
+		if 0 < c && c < k {
+			split = 'a' + byte(i)
+			break
+		}
+	}
+	if split == 0 {
+		return len(s)
+	}
+	for _, subStr := range strings.Split(s, string(split)) {
+		res = max(res, longestSubstring1(subStr, k))
+	}
+	return res
+}
+```