Add solution 6、8、12

Author: halfrost

README.md

@@ -0,0 +1,26 @@
+package leetcode
+
+func convert(s string, numRows int) string {
+	matrix, down, up := make([][]byte, numRows, numRows), 0, numRows-2
+	for i := 0; i != len(s); {
+		if down != numRows {
+			matrix[down] = append(matrix[down], byte(s[i]))
+			down++
+			i++
+		} else if up > 0 {
+			matrix[up] = append(matrix[up], byte(s[i]))
+			up--
+			i++
+		} else {
+			up = numRows - 2
+			down = 0
+		}
+	}
+	solution := make([]byte, 0, len(s))
+	for _, row := range matrix {
+		for _, item := range row {
+			solution = append(solution, item)
+		}
+	}
+	return string(solution)
+}

leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion.go

@@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question6 struct {
+	para6
+	ans6
+}
+
+// para 是参数
+// one 代表第一个参数
+type para6 struct {
+	s       string
+	numRows int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans6 struct {
+	one string
+}
+
+func Test_Problem6(t *testing.T) {
+
+	qs := []question6{
+
+		{
+			para6{"PAYPALISHIRING", 3},
+			ans6{"PAHNAPLSIIGYIR"},
+		},
+
+		{
+			para6{"PAYPALISHIRING", 4},
+			ans6{"PINALSIGYAHRPI"},
+		},
+
+		{
+			para6{"A", 1},
+			ans6{"A"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 6------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans6, q.para6
+		fmt.Printf("【input】:%v       【output】:%v\n", p, convert(p.s, p.numRows))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion_test.go

@@ -0,0 +1,107 @@
+# [6. ZigZag Conversion](https://leetcode.com/problems/zigzag-conversion/)
+
+
+## 题目
+
+The string `"PAYPALISHIRING"` is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
+
+```
+P   A   H   N
+A P L S I I G
+Y   I   R
+```
+
+And then read line by line: `"PAHNAPLSIIGYIR"`
+
+Write the code that will take a string and make this conversion given a number of rows:
+
+```
+string convert(string s, int numRows);
+```
+
+**Example 1:**
+
+```
+Input: s = "PAYPALISHIRING", numRows = 3
+Output: "PAHNAPLSIIGYIR"
+```
+
+**Example 2:**
+
+```
+Input: s = "PAYPALISHIRING", numRows = 4
+Output: "PINALSIGYAHRPI"
+Explanation:
+P     I    N
+A   L S  I G
+Y A   H R
+P     I
+```
+
+**Example 3:**
+
+```
+Input: s = "A", numRows = 1
+Output: "A"
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 1000`
+- `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
+- `1 <= numRows <= 1000`
+
+## 题目大意
+
+将一个给定字符串 `s` 根据给定的行数 `numRows` ，以从上往下、从左到右进行 Z 字形排列。
+
+比如输入字符串为 `"PAYPALISHIRING"` 行数为 3 时，排列如下：
+
+```go
+P   A   H   N
+A P L S I I G
+Y   I   R
+```
+
+之后，你的输出需要从左往右逐行读取，产生出一个新的字符串，比如：`"PAHNAPLSIIGYIR"`。
+
+请你实现这个将字符串进行指定行数变换的函数：
+
+```go
+string convert(string s, int numRows);
+```
+
+## 解题思路
+
+- 这一题没有什么算法思想，考察的是对程序控制的能力。用 2 个变量保存方向，当垂直输出的行数达到了规定的目标行数以后，需要从下往上转折到第一行，循环中控制好方向ji
+
+## 代码
+
+```go
+package leetcode
+
+func convert(s string, numRows int) string {
+	matrix, down, up := make([][]byte, numRows, numRows), 0, numRows-2
+	for i := 0; i != len(s); {
+		if down != numRows {
+			matrix[down] = append(matrix[down], byte(s[i]))
+			down++
+			i++
+		} else if up > 0 {
+			matrix[up] = append(matrix[up], byte(s[i]))
+			up--
+			i++
+		} else {
+			up = numRows - 2
+			down = 0
+		}
+	}
+	solution := make([]byte, 0, len(s))
+	for _, row := range matrix {
+		for _, item := range row {
+			solution = append(solution, item)
+		}
+	}
+	return string(solution)
+}
+```

leetcode/0006.ZigZag-Conversion/README.md

@@ -0,0 +1,56 @@
+package leetcode
+
+func myAtoi(s string) int {
+	maxInt, signAllowed, whitespaceAllowed, sign, digits := int64(2<<30), true, true, 1, []int{}
+	for _, c := range s {
+		if c == ' ' && whitespaceAllowed {
+			continue
+		}
+		if signAllowed {
+			if c == '+' {
+				signAllowed = false
+				whitespaceAllowed = false
+				continue
+			} else if c == '-' {
+				sign = -1
+				signAllowed = false
+				whitespaceAllowed = false
+				continue
+			}
+		}
+		if c < '0' || c > '9' {
+			break
+		}
+		whitespaceAllowed, signAllowed = false, false
+		digits = append(digits, int(c-48))
+	}
+	var num, place int64
+	place, num = 1, 0
+	lastLeading0Index := -1
+	for i, d := range digits {
+		if d == 0 {
+			lastLeading0Index = i
+		} else {
+			break
+		}
+	}
+	if lastLeading0Index > -1 {
+		digits = digits[lastLeading0Index+1:]
+	}
+	var rtnMax int64
+	if sign > 0 {
+		rtnMax = maxInt - 1
+	} else {
+		rtnMax = maxInt
+	}
+	digitsCount := len(digits)
+	for i := digitsCount - 1; i >= 0; i-- {
+		num += int64(digits[i]) * place
+		place *= 10
+		if digitsCount-i > 10 || num > rtnMax {
+			return int(int64(sign) * rtnMax)
+		}
+	}
+	num *= int64(sign)
+	return int(num)
+}