Merge branch 'master' into janetyu

Author: Janetyu

README.md

@@ -5,20 +5,23 @@ 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
+	var bitSet [256]bool
 	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
+		// 右侧字符对应的bitSet被标记true，说明此字符在X位置重复，需要左侧向前移动，直到将X标记为false
+		if bitSet[s[right]] {
+			bitSet[s[left]] = false
 			left++
+		} else {
+			bitSet[s[right]] = true
+			right++
+		}
+		if result < right-left {
+			result = right - left
+		}
+		if left+result >= len(s) || right >= len(s) {
+			break
 		}
-		result = max(result, right-left)
 	}
 	return result
 }

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

@@ -0,0 +1,17 @@
+package leetcode
+
+func generate(numRows int) [][]int {
+	result := [][]int{}
+	for i := 0; i < numRows; i++ {
+		row := []int{}
+		for j := 0; j < i+1; j++ {
+			if j == 0 || j == i {
+				row = append(row, 1)
+			} else if i > 1 {
+				row = append(row, result[i-1][j-1]+result[i-1][j])
+			}
+		}
+		result = append(result, row)
+	}
+	return result
+}

leetcode/0118.Pascals-Triangle/118. Pascal's Triangle.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question118 struct {
+	para118
+	ans118
+}
+
+// para 是参数
+// one 代表第一个参数
+type para118 struct {
+	numRows int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans118 struct {
+	one [][]int
+}
+
+func Test_Problem118(t *testing.T) {
+
+	qs := []question118{
+
+		question118{
+			para118{2},
+			ans118{[][]int{{1}, {1, 1}}},
+		},
+
+		question118{
+			para118{5},
+			ans118{[][]int{{1}, {1, 1}, {1, 2, 1}, {1, 3, 3, 1}, {1, 4, 6, 4, 1}}},
+		},
+
+		question118{
+			para118{10},
+			ans118{[][]int{{1}, {1, 1}, {1, 2, 1}, {1, 3, 3, 1}, {1, 4, 6, 4, 1}, {1, 5, 10, 10, 5, 1}, {1, 6, 15, 20, 15, 6, 1}, {1, 7, 21, 35, 35, 21, 7, 1}, {1, 8, 28, 56, 70, 56, 28, 8, 1}, {1, 9, 36, 84, 126, 126, 84, 36, 9, 1}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 118------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans118, q.para118
+		fmt.Printf("【input】:%v    【output】:%v\n", p, generate(p.numRows))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0118.Pascals-Triangle/118. Pascal's Triangle_test.go

@@ -0,0 +1,60 @@
+# [118. Pascal's Triangle](https://leetcode.com/problems/pascals-triangle/)
+
+
+## 题目
+
+Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.
+
+![](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif)
+
+**Note:** In Pascal's triangle, each number is the sum of the two numbers directly above it.
+
+**Example:**
+
+```
+Input: 5
+Output:
+[
+     [1],
+    [1,1],
+   [1,2,1],
+  [1,3,3,1],
+ [1,4,6,4,1]
+]
+```
+
+## 题目大意
+
+给定一个非负整数 numRows，生成杨辉三角的前 numRows 行。在杨辉三角中，每个数是它左上方和右上方的数的和。
+
+
+## 解题思路
+
+- 给定一个 n，要求打印杨辉三角的前 n 行。
+- 简单题。按照杨辉三角的生成规则循环打印即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func generate(numRows int) [][]int {
+	result := [][]int{}
+	for i := 0; i < numRows; i++ {
+		row := []int{}
+		for j := 0; j < i+1; j++ {
+			if j == 0 || j == i {
+				row = append(row, 1)
+			} else if i > 1 {
+				row = append(row, result[i-1][j-1]+result[i-1][j])
+			}
+		}
+		result = append(result, row)
+	}
+	return result
+}
+
+```
+

leetcode/0118.Pascals-Triangle/README.md

@@ -60,20 +60,23 @@ 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
+	var bitSet [256]bool
 	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
+		// 右侧字符对应的 bitSet 被标记 true，说明此字符在 X 位置重复，需要左侧向前移动，直到将X标记为 false
+		if bitSet[s[right]] {
+			bitSet[s[left]] = false
 			left++
+		} else {
+			bitSet[s[right]] = true
+			right++
+		}
+		if result < right-left {
+			result = right - left
+		}
+		if left+result >= len(s) || right >= len(s) {
+			break
 		}
-		result = max(result, right-left)
 	}
 	return result
 }

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

@@ -0,0 +1,60 @@
+# [118. Pascal's Triangle](https://leetcode.com/problems/pascals-triangle/)
+
+
+## 题目
+
+Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.
+
+![](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif)
+
+**Note:** In Pascal's triangle, each number is the sum of the two numbers directly above it.
+
+**Example:**
+
+```
+Input: 5
+Output:
+[
+     [1],
+    [1,1],
+   [1,2,1],
+  [1,3,3,1],
+ [1,4,6,4,1]
+]
+```
+
+## 题目大意
+
+给定一个非负整数 numRows，生成杨辉三角的前 numRows 行。在杨辉三角中，每个数是它左上方和右上方的数的和。
+
+
+## 解题思路
+
+- 给定一个 n，要求打印杨辉三角的前 n 行。
+- 简单题。按照杨辉三角的生成规则循环打印即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func generate(numRows int) [][]int {
+	result := [][]int{}
+	for i := 0; i < numRows; i++ {
+		row := []int{}
+		for j := 0; j < i+1; j++ {
+			if j == 0 || j == i {
+				row = append(row, 1)
+			} else if i > 1 {
+				row = append(row, result[i-1][j-1]+result[i-1][j])
+			}
+		}
+		result = append(result, row)
+	}
+	return result
+}
+
+```
+

website/content/ChapterFour/0118.Pascals-Triangle.md

@@ -125,6 +125,7 @@ headless: true
     - [0112.Path-Sum]({{< relref "/ChapterFour/0112.Path-Sum.md" >}})
     - [0113.Path-Sum-II]({{< relref "/ChapterFour/0113.Path-Sum-II.md" >}})
     - [0114.Flatten-Binary-Tree-to-Linked-List]({{< relref "/ChapterFour/0114.Flatten-Binary-Tree-to-Linked-List.md" >}})
+    - [0118.Pascals-Triangle]({{< relref "/ChapterFour/0118.Pascals-Triangle.md" >}})
     - [0120.Triangle]({{< relref "/ChapterFour/0120.Triangle.md" >}})
     - [0121.Best-Time-to-Buy-and-Sell-Stock]({{< relref "/ChapterFour/0121.Best-Time-to-Buy-and-Sell-Stock.md" >}})
     - [0122.Best-Time-to-Buy-and-Sell-Stock-II]({{< relref "/ChapterFour/0122.Best-Time-to-Buy-and-Sell-Stock-II.md" >}})