Merge pull request #45 from halfrost/add_hugo

Optimized solution 3、39 & update README

README.md

@@ -8,7 +8,7 @@
 </p>
 
 <p align='center'>
-<a href="https://leetcode.com/halfrost/"><img src="https://img.shields.io/badge/@halfrost-9650-yellow.svg">
+<a href="https://leetcode.com/halfrost/"><img src="https://img.shields.io/badge/@halfrost-8751-yellow.svg">
 <img src="https://img.shields.io/badge/build-passing-brightgreen.svg">
 <img src="https://img.shields.io/badge/language-Golang-abcdef.svg">
 <a href="https://halfrost.com"><img src="https://img.shields.io/badge/Blog-Halfrost--Field-80d4f9.svg?style=flat"></a>
@@ -26,13 +26,15 @@
 </p>
 
 <p align='center'>
-支持 Progressive Web Apps 的题解电子书《LeetCode Cookbook》在线阅读 <a href="https://books.halfrost.com/leetcode/" rel="nofollow">地址</a>
+支持 Progressive Web Apps 的题解电子书《LeetCode Cookbook》 <a href="https://books.halfrost.com/leetcode/" rel="nofollow">Online Reading</a>
 </p>
 
 ## Data Structures
 
 > 标识了 ✅ 的专题是完成所有题目了的，没有标识的是还没有做完所有题目的
 
+<a href="https://books.halfrost.com/leetcode/"><img src="./website/static/logo.png" alt="logo" height="550" align="right" /></a>
+
 * [Array](#array)
 * [String](#string)
 * [Two Pointers ✅](#two-pointers)

leetcode/0003.Longest-Substring-Without-Repeating-Characters/3. Longest Substring Without Repeating Characters.go

@@ -1,12 +1,35 @@
 package leetcode
 
+// 解法一 位图
 func lengthOfLongestSubstring(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	// 扩展 ASCII 码的位图表示（BitSet），共有 256 位
+	var bitSet [256]uint8
+	result, left, right := 0, 0, 0
+	for left < len(s) {
+		if right < len(s) && bitSet[s[right]] == 0 {
+			// 标记对应的 ASCII 码为 1
+			bitSet[s[right]] = 1
+			right++
+		} else {
+			// 标记对应的 ASCII 码为 0
+			bitSet[s[left]] = 0
+			left++
+		}
+		result = max(result, right-left)
+	}
+	return result
+}
+
+// 解法二 滑动窗口
+func lengthOfLongestSubstring_(s string) int {
 	if len(s) == 0 {
 		return 0
 	}
 	var freq [256]int
-	result := 0
-	left, right := 0, -1
+	result, left, right := 0, 0, -1
 
 	for left < len(s) {
 		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {

leetcode/0003.Longest-Substring-Without-Repeating-Characters/3. Longest Substring Without Repeating Characters_test.go

@@ -51,7 +51,7 @@ func Test_Problem3(t *testing.T) {
 
 	for _, q := range qs {
 		_, p := q.ans3, q.para3
-		fmt.Printf("【input】:%v       【output】:%v\n", p, lengthOfLongestSubstring(p.s))
+		fmt.Printf("【input】:%v       【output】:%v\n", p, lengthOfLongestSubstring_(p.s))
 	}
 	fmt.Printf("\n\n\n")
 }

leetcode/0039.Combination-Sum/39. Combination Sum.go

@@ -22,6 +22,9 @@ func findcombinationSum(nums []int, target, index int, c []int, res *[][]int) {
 		return
 	}
 	for i := index; i < len(nums); i++ {
+		if nums[i] > target { // 这里可以剪枝优化
+			break
+		}
 		c = append(c, nums[i])
 		findcombinationSum(nums, target-nums[i], i, c, res) // 注意这里迭代的时候 index 依旧不变，因为一个元素可以取多次
 		c = c[:len(c)-1]

website/content/ChapterFour/0003.Longest-Substring-Without-Repeating-Characters.md

@@ -55,13 +55,36 @@ Explanation: The answer is "wke", with the length of 3.
 
 package leetcode
 
+// 解法一 位图
 func lengthOfLongestSubstring(s string) int {
+	if len(s) == 0 {
+		return 0
+	}
+	// 扩展 ASCII 码的位图表示（BitSet），共有 256 位
+	var bitSet [256]uint8
+	result, left, right := 0, 0, 0
+	for left < len(s) {
+		if right < len(s) && bitSet[s[right]] == 0 {
+			// 标记对应的 ASCII 码为 1
+			bitSet[s[right]] = 1
+			right++
+		} else {
+			// 标记对应的 ASCII 码为 0
+			bitSet[s[left]] = 0
+			left++
+		}
+		result = max(result, right-left)
+	}
+	return result
+}
+
+// 解法二 滑动窗口
+func lengthOfLongestSubstring_(s string) int {
 	if len(s) == 0 {
 		return 0
 	}
 	var freq [256]int
-	result := 0
-	left, right := 0, -1
+	result, left, right := 0, 0, -1
 
 	for left < len(s) {
 		if right+1 < len(s) && freq[s[right+1]-'a'] == 0 {
@@ -76,6 +99,14 @@ func lengthOfLongestSubstring(s string) int {
 	return result
 }
 
+func max(a int, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+
 ```
 
 

website/content/ChapterFour/0039.Combination-Sum.md

@@ -75,11 +75,13 @@ func findcombinationSum(nums []int, target, index int, c []int, res *[][]int) {
 		return
 	}
 	for i := index; i < len(nums); i++ {
+		if nums[i] > target { // 这里可以剪枝优化
+			break
+		}
 		c = append(c, nums[i])
 		findcombinationSum(nums, target-nums[i], i, c, res) // 注意这里迭代的时候 index 依旧不变，因为一个元素可以取多次
 		c = c[:len(c)-1]
 	}
 }
 
-
 ```