Add solution 43、45、73、227

Author: halfrost

README.md

@@ -0,0 +1,62 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question8 struct {
+	para8
+	ans8
+}
+
+// para 是参数
+// one 代表第一个参数
+type para8 struct {
+	one string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans8 struct {
+	one int
+}
+
+func Test_Problem8(t *testing.T) {
+
+	qs := []question8{
+
+		{
+			para8{"42"},
+			ans8{42},
+		},
+
+		{
+			para8{"   -42"},
+			ans8{-42},
+		},
+
+		{
+			para8{"4193 with words"},
+			ans8{4193},
+		},
+
+		{
+			para8{"words and 987"},
+			ans8{0},
+		},
+
+		{
+			para8{"-91283472332"},
+			ans8{-2147483648},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 8------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans8, q.para8
+		fmt.Printf("【input】:%v    【output】:%v\n", p.one, myAtoi(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0008.String-to-Integer-atoi/8. String to Integer (atoi)_test.go

@@ -0,0 +1,25 @@
+package leetcode
+
+func multiply(num1 string, num2 string) string {
+	if num1 == "0" || num2 == "0" {
+		return "0"
+	}
+	b1, b2, tmp := []byte(num1), []byte(num2), make([]int, len(num1)+len(num2))
+	for i := 0; i < len(b1); i++ {
+		for j := 0; j < len(b2); j++ {
+			tmp[i+j+1] += int(b1[i]-'0') * int(b2[j]-'0')
+		}
+	}
+	for i := len(tmp) - 1; i > 0; i-- {
+		tmp[i-1] += tmp[i] / 10
+		tmp[i] = tmp[i] % 10
+	}
+	if tmp[0] == 0 {
+		tmp = tmp[1:]
+	}
+	res := make([]byte, len(tmp))
+	for i := 0; i < len(tmp); i++ {
+		res[i] = '0' + byte(tmp[i])
+	}
+	return string(res)
+}

leetcode/0008.String-to-Integer-atoi/8. String to Integer (atoi).go renamed to leetcode/0008.String-to-Integer-atoi/8. String to Integer atoi.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question43 struct {
+	para43
+	ans43
+}
+
+// para 是参数
+// one 代表第一个参数
+type para43 struct {
+	num1 string
+	num2 string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans43 struct {
+	one string
+}
+
+func Test_Problem43(t *testing.T) {
+
+	qs := []question43{
+
+		{
+			para43{"2", "3"},
+			ans43{"6"},
+		},
+
+		{
+			para43{"123", "456"},
+			ans43{"56088"},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 43------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans43, q.para43
+		fmt.Printf("【input】:%v       【output】:%v\n", p, multiply(p.num1, p.num2))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0008.String-to-Integer-atoi/8. String to Integer atoi_test.go

@@ -0,0 +1,66 @@
+# [43. Multiply Strings](https://leetcode.com/problems/multiply-strings/)
+
+
+## 题目
+
+Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.
+
+**Note:** You must not use any built-in BigInteger library or convert the inputs to integer directly.
+
+**Example 1:**
+
+```
+Input: num1 = "2", num2 = "3"
+Output: "6"
+```
+
+**Example 2:**
+
+```
+Input: num1 = "123", num2 = "456"
+Output: "56088"
+```
+
+**Constraints:**
+
+- `1 <= num1.length, num2.length <= 200`
+- `num1` and `num2` consist of digits only.
+- Both `num1` and `num2` do not contain any leading zero, except the number `0` itself.
+
+## 题目大意
+
+给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。
+
+## 解题思路
+
+- 用数组模拟乘法。创建一个数组长度为 `len(num1) + len(num2)` 的数组用于存储乘积。对于任意 `0 ≤ i < len(num1)`，`0 ≤ j < len(num2)`，`num1[i] * num2[j]` 的结果位于 `tmp[i+j+1]`，如果 `tmp[i+j+1]≥10`，则将进位部分加到 `tmp[i+j]`。最后，将数组 `tmp` 转成字符串，如果最高位是 0 则舍弃最高位。
+
+## 代码
+
+```go
+package leetcode
+
+func multiply(num1 string, num2 string) string {
+	if num1 == "0" || num2 == "0" {
+		return "0"
+	}
+	b1, b2, tmp := []byte(num1), []byte(num2), make([]int, len(num1)+len(num2))
+	for i := 0; i < len(b1); i++ {
+		for j := 0; j < len(b2); j++ {
+			tmp[i+j+1] += int(b1[i]-'0') * int(b2[j]-'0')
+		}
+	}
+	for i := len(tmp) - 1; i > 0; i-- {
+		tmp[i-1] += tmp[i] / 10
+		tmp[i] = tmp[i] % 10
+	}
+	if tmp[0] == 0 {
+		tmp = tmp[1:]
+	}
+	res := make([]byte, len(tmp))
+	for i := 0; i < len(tmp); i++ {
+		res[i] = '0' + byte(tmp[i])
+	}
+	return string(res)
+}
+```

leetcode/0043.Multiply-Strings/43. Multiply Strings.go

@@ -0,0 +1,21 @@
+package leetcode
+
+func jump(nums []int) int {
+	if len(nums) == 1 {
+		return 0
+	}
+	needChoose, canReach, step := 0, 0, 0
+	for i, x := range nums {
+		if i+x > canReach {
+			canReach = i + x
+			if canReach >= len(nums)-1 {
+				return step + 1
+			}
+		}
+		if i == needChoose {
+			needChoose = canReach
+			step++
+		}
+	}
+	return step
+}

leetcode/0043.Multiply-Strings/43. Multiply Strings_test.go

@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question45 struct {
+	para45
+	ans45
+}
+
+// para 是参数
+// one 代表第一个参数
+type para45 struct {
+	nums []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans45 struct {
+	one int
+}
+
+func Test_Problem45(t *testing.T) {
+
+	qs := []question45{
+
+		{
+			para45{[]int{2, 3, 1, 1, 4}},
+			ans45{2},
+		},
+
+		{
+			para45{[]int{2, 3, 0, 1, 4}},
+			ans45{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 45------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans45, q.para45
+		fmt.Printf("【input】:%v       【output】:%v\n", p, jump(p.nums))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0043.Multiply-Strings/README.md

@@ -0,0 +1,67 @@
+# [45. Jump Game II](https://leetcode.com/problems/jump-game-ii/)
+
+
+## 题目
+
+Given an array of non-negative integers `nums`, you are initially positioned at the first index of the array.
+
+Each element in the array represents your maximum jump length at that position.
+
+Your goal is to reach the last index in the minimum number of jumps.
+
+You can assume that you can always reach the last index.
+
+**Example 1:**
+
+```
+Input: nums = [2,3,1,1,4]
+Output: 2
+Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,3,0,1,4]
+Output: 2
+```
+
+**Constraints:**
+
+- `1 <= nums.length <= 1000`
+- `0 <= nums[i] <= 10^5`
+
+## 题目大意
+
+给定一个非负整数数组，你最初位于数组的第一个位置。数组中的每个元素代表你在该位置可以跳跃的最大长度。你的目标是使用最少的跳跃次数到达数组的最后一个位置。
+
+## 解题思路
+
+- 要求找到最少跳跃次数，顺理成章的会想到用贪心算法解题。扫描步数数组，维护当前能够到达最大下标的位置，记为能到达的最远边界，如果扫描过程中到达了最远边界，更新边界并将跳跃次数 + 1。
+- 扫描数组的时候，其实不需要扫描最后一个元素，因为在跳到最后一个元素之前，能到达的最远边界一定大于等于最后一个元素的位置，不然就跳不到最后一个元素，到达不了终点了；如果遍历到最后一个元素，说明边界正好为最后一个位置，最终跳跃次数直接 + 1 即可，也不需要访问最后一个元素。
+
+## 代码
+
+```go
+package leetcode
+
+func jump(nums []int) int {
+	if len(nums) == 1 {
+		return 0
+	}
+	needChoose, canReach, step := 0, 0, 0
+	for i, x := range nums {
+		if i+x > canReach {
+			canReach = i + x
+			if canReach >= len(nums)-1 {
+				return step + 1
+			}
+		}
+		if i == needChoose {
+			needChoose = canReach
+			step++
+		}
+	}
+	return step
+}
+```

leetcode/0045.Jump-Game-II/45. Jump Game II.go

@@ -0,0 +1,54 @@
+package leetcode
+
+func setZeroes(matrix [][]int) {
+	if len(matrix) == 0 || len(matrix[0]) == 0 {
+		return
+	}
+	isFirstRowExistZero, isFirstColExistZero := false, false
+	for i := 0; i < len(matrix); i++ {
+		if matrix[i][0] == 0 {
+			isFirstColExistZero = true
+			break
+		}
+	}
+	for j := 0; j < len(matrix[0]); j++ {
+		if matrix[0][j] == 0 {
+			isFirstRowExistZero = true
+			break
+		}
+	}
+	for i := 1; i < len(matrix); i++ {
+		for j := 1; j < len(matrix[0]); j++ {
+			if matrix[i][j] == 0 {
+				matrix[i][0] = 0
+				matrix[0][j] = 0
+			}
+		}
+	}
+	// 处理[1:]行全部置 0
+	for i := 1; i < len(matrix); i++ {
+		if matrix[i][0] == 0 {
+			for j := 1; j < len(matrix[0]); j++ {
+				matrix[i][j] = 0
+			}
+		}
+	}
+	// 处理[1:]列全部置 0
+	for j := 1; j < len(matrix[0]); j++ {
+		if matrix[0][j] == 0 {
+			for i := 1; i < len(matrix); i++ {
+				matrix[i][j] = 0
+			}
+		}
+	}
+	if isFirstRowExistZero {
+		for j := 0; j < len(matrix[0]); j++ {
+			matrix[0][j] = 0
+		}
+	}
+	if isFirstColExistZero {
+		for i := 0; i < len(matrix); i++ {
+			matrix[i][0] = 0
+		}
+	}
+}