添加 problem 438

Algorithms/0438. Find All Anagrams in a String/438. Find All Anagrams in a String.go

@@ -0,0 +1,33 @@
+package leetcode
+
+func findAnagrams(s string, p string) []int {
+	var freq [256]int
+	result := []int{}
+	if len(s) == 0 || len(s) < len(p) {
+		return result
+	}
+	for i := 0; i < len(p); i++ {
+		freq[p[i]-'a']++
+	}
+	left, right, count := 0, 0, len(p)
+
+	for right < len(s) {
+		if freq[s[right]-'a'] >= 1 {
+			count--
+		}
+		freq[s[right]-'a']--
+		right++
+		if count == 0 {
+			result = append(result, left)
+		}
+		if right-left == len(p) {
+			if freq[s[left]-'a'] >= 0 {
+				count++
+			}
+			freq[s[left]-'a']++
+			left++
+		}
+
+	}
+	return result
+}

Algorithms/0438. Find All Anagrams in a String/438. Find All Anagrams in a String_test.go

@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question438 struct {
+	para438
+	ans438
+}
+
+// para 是参数
+// one 代表第一个参数
+type para438 struct {
+	s string
+	p string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans438 struct {
+	one []int
+}
+
+func Test_Problem438(t *testing.T) {
+
+	qs := []question438{
+
+		question438{
+			para438{"abab", "ab"},
+			ans438{[]int{0, 1, 2}},
+		},
+
+		question438{
+			para438{"cbaebabacd", "abc"},
+			ans438{[]int{0, 6}},
+		},
+
+		question438{
+			para438{"", "abc"},
+			ans438{[]int{}},
+		},
+
+		question438{
+			para438{"abacbabc", "abc"},
+			ans438{[]int{1, 2, 3, 5}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 438------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans438, q.para438
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findAnagrams(p.s, p.p))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/0438. Find All Anagrams in a String/README.md

@@ -0,0 +1,61 @@
+# [438. Find All Anagrams in a String](https://leetcode.com/problems/find-all-anagrams-in-a-string/)
+
+## 题目
+
+Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
+
+Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
+
+The order of output does not matter.
+
+
+Example 1:
+
+```c
+Input:
+s: "cbaebabacd" p: "abc"
+
+Output:
+[0, 6]
+
+Explanation:
+The substring with start index = 0 is "cba", which is an anagram of "abc".
+The substring with start index = 6 is "bac", which is an anagram of "abc".
+```
+
+Example 2:
+
+```c
+Input:
+s: "abab" p: "ab"
+
+Output:
+[0, 1, 2]
+
+Explanation:
+The substring with start index = 0 is "ab", which is an anagram of "ab".
+The substring with start index = 1 is "ba", which is an anagram of "ab".
+The substring with start index = 2 is "ab", which is an anagram of "ab".
+```
+
+## 题目大意
+
+给定一个字符串 s 和一个非空字符串 p，找出 s 中的所有是 p 的 Anagrams 字符串的子串，返回这些子串的起始索引。Anagrams 的意思是和一个字符串的所有字符都一样，只是排列组合不同。
+
+## 解题思路
+
+这道题是一道考“滑动窗口”的题目。和第 3 题，第 76 题，第 567 题类似的。解法也是用 freq[256] 记录每个字符的出现的频次次数。滑动窗口左边界往右滑动的时候，划过去的元素释放次数(即次数 ++)，滑动窗口右边界往右滑动的时候，划过去的元素消耗次数(即次数 \-\-)。右边界和左边界相差 len(p) 的时候，需要判断每个元素是否都用过一遍了。具体做法是每经过一个符合规范的元素，count 就 \-\-，count 初始值是 len(p)，当每个元素都符合规范的时候，右边界和左边界相差 len(p) 的时候，count 也会等于 0 。当区间内有不符合规范的元素(freq < 0 或者是不存在的元素)，那么当区间达到 len(p) 的时候，count 无法减少到 0，区间右移动的时候，左边界又会开始 count ++，只有当左边界移出了这些不合规范的元素以后，才可能出现 count = 0 的情况，即找到 Anagrams 的情况。
+
+
+
+
+
+
+
+
+
+
+
+
+
+