目录

1. 全排列  🌟🌟

2. 分数到小数  🌟🌟

3. 删除排序链表中的重复元素 II  🌟🌟

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---

1. 全排列

给定一个 没有重复 数字的序列，返回其所有可能的全排列。

示例:

输入: [1,2,3]
输出: [[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]]

以下程序实现了这一功能，请你填补空白处内容：

···
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    vector<vector<int>> permute(vector<int> &nums)
    {
        vector<vector<int>> res;
        vector<bool> used(nums.size());
        dfs(nums, used, res);
        return res;
    }
private:
    vector<int> stack;
    void dfs(vector<int> &nums, vector<bool> &used, vector<vector<int>> &res)
    {
        if (stack.size() == nums.size())
        {
            res.push_back(stack);
        }
        else
        {
            for (int i = 0; i < nums.size(); i++)
            {
                if (!used[i])
                {
                    _______________;
                }
            }
        }
    }
};
···

出处：

https://edu.csdn.net/practice/27810770

代码：

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    vector<vector<int>> permute(vector<int> &nums)
    {
        vector<vector<int>> res;
        vector<bool> used(nums.size());
        dfs(nums, used, res);
        return res;
    }
private:
    vector<int> stack;
    void dfs(vector<int> &nums, vector<bool> &used, vector<vector<int>> &res)
    {
        if (stack.size() == nums.size())
        {
            res.push_back(stack);
        }
        else
        {
            for (int i = 0; i < nums.size(); i++)
            {
                if (!used[i])
                {
                    used[i] = true;
                    stack.push_back(nums[i]);
                    dfs(nums, used, res);
                    stack.pop_back();
                    used[i] = false;
                }
            }
        }
    }
};

string Vector2dToString(vector<vector<int>> vec2d, string sep = ",")
{
    stringstream ss;
    ss << "[";
    for (int i = 0; i < vec2d.size(); ++i) {
        ss << "[";
        copy(vec2d[i].begin(), vec2d[i].end(), ostream_iterator<int>(ss, sep.c_str()));
		ss.seekp(-(int)sep.size(), ios_base::end);
        ss << "]" << sep;
    }
    ss.seekp(-(int)sep.size(), ios_base::end);
    ss << "]";
    return ss.str();
}

int main()
{
	Solution s;
	vector<int> nums = {1,2,3};
	cout << Vector2dToString(s.permute(nums)) << endl;
	
	return 0;
} 

输出：

[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

---

2. 分数到小数

给定两个整数，分别表示分数的分子 numerator 和分母 denominator，以 字符串形式返回小数 。

如果小数部分为循环小数，则将循环的部分括在括号内。

如果存在多个答案，只需返回 任意一个 。

对于所有给定的输入，保证 答案字符串的长度小于 104 。

示例 1：

输入：numerator = 1, denominator = 2
输出："0.5"

示例 2：

输入：numerator = 2, denominator = 1
输出："2"

示例 3：

输入：numerator = 2, denominator = 3
输出："0.(6)"

示例 4：

输入：numerator = 4, denominator = 333
输出："0.(012)"

示例 5：

输入：numerator = 1, denominator = 5
输出："0.2"

提示：

• -2^31 <= numerator, denominator <= 2^31 - 1
• denominator != 0

出处：

https://edu.csdn.net/practice/27810771

代码：

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    string fractionToDecimal(int numerator, int denominator)
    {
        if (denominator == 0)
            return "";
        string ans;
        if ((numerator > 0 && denominator < 0) || (numerator < 0 && denominator > 0))
            ans = "-";
        long long num = abs(static_cast<long long>(numerator));
        long long den = abs(static_cast<long long>(denominator));
        long long quo = num / den;
        long long rem = num % den;
        ans = ans + to_string(quo);
        if (!rem)
            return ans;
        ans += ".";
        unordered_map<long long, int> u_m;
        string s = "";
        int pos = 0;
        while (rem)
        {
            if (u_m.find(rem) != u_m.end())
            {
                s.insert(u_m[rem], "(");
                s += ")";
                return ans + s;
            }
            u_m[rem] = pos++;
            s += to_string((rem * 10) / den);
            rem = (rem * 10) % den;
        }
        return ans + s;
    }
};

int main()
{
	Solution s;
	cout << s.fractionToDecimal(1, 2) << endl;
	cout << s.fractionToDecimal(2, 1) << endl;
	cout << s.fractionToDecimal(2, 3) << endl;
	cout << s.fractionToDecimal(4, 333) << endl;
	cout << s.fractionToDecimal(1, 5) << endl;
	
	return 0;
} 

输出：

0.5
2
0.(6)
0.(012)
0.2

---

3. 删除排序链表中的重复元素 II （C语言）

存在一个按升序排列的链表，给你这个链表的头节点 head ，请你删除链表中所有存在数字重复情况的节点，只保留原始链表中 没有重复出现 的数字。

返回同样按升序排列的结果链表。

示例 1：

输入：head = [1,2,3,3,4,4,5]
输出：[1,2,5]

示例 2：

输入：head = [1,1,1,2,3]
输出：[2,3]

提示：

• 链表中节点数目在范围 [0, 300] 内
• -100 <= Node.val <= 100
• 题目数据保证链表已经按升序排列

以下程序实现了这一功能，请你填补空白处内容：

```c++
#include <stdio.h>
#include <stdlib.h>
struct ListNode
{
    int val;
    struct ListNode *next;
};
struct ListNode *deleteDuplicates(struct ListNode *head)
{
    struct ListNode dummy;
    struct ListNode *p, *q, *prev;
    prev = &dummy;
    dummy.next = head;
    p = q = head;
    while (p != NULL)
    {
        _____________________;
    }
    return dummy.next;
}
int main(int argc, char **argv)
{
    int i;
    struct ListNode *head = NULL;
    struct ListNode *prev = NULL;
    struct ListNode *p;
    for (i = 0; i < argc - 1; i++)
    {
        p = malloc(sizeof(*p));
        p->val = atoi(argv[i + 1]);
        p->next = NULL;
        if (head == NULL)
        {
            head = p;
            prev = head;
        }
        else
        {
            prev->next = p;
            prev = p;
        }
    }
    p = deleteDuplicates(head);
    while (p != NULL)
    {
        printf("%d ", p->val);
        p = p->next;
    }
    printf("\n");
    return 0;
}
```

出处：

https://edu.csdn.net/practice/27810772

代码： (C语言)

#include <stdio.h>
#include <stdlib.h>
struct ListNode
{
    int val;
    struct ListNode *next;
};
struct ListNode *deleteDuplicates(struct ListNode *head)
{
    struct ListNode dummy;
    struct ListNode *p, *q, *prev;
    prev = &dummy;
    dummy.next = head;
    p = q = head;
    while (p != NULL)
    {
        while (q != NULL && q->val == p->val)
		{
		    q = q->next;
		}
		if (p->next == q)
		{
		    prev = p;
		}
		else
		{
		    prev->next = q;
		}
		p = q;
    }
    return dummy.next;
}

int main()
{
    int i;
    struct ListNode *head = NULL;
    struct ListNode *prev = NULL;
    struct ListNode *p;
    int nums[] = {1,2,3,3,4,4,5};
    int count = sizeof(nums) / sizeof(nums[0]);
    for (i = 0; i < count; i++)
    {
        p = (ListNode*)malloc(sizeof(*p));
        p->val = nums[i];
        p->next = NULL;
        if (head == NULL)
        {
            head = p;
            prev = head;
        }
        else
        {
            prev->next = p;
            prev = p;
        }
    }
    p = deleteDuplicates(head);
    while (p != NULL)
    {
        printf("%d->", p->val);
        p = p->next;
    }
    printf("null\n");
    return 0;
}

输出：

1->2->5->null

单链表的创建、打印写成函数形式：

#include <stdio.h>
#include <stdlib.h>
#define Test(nums) {\
		count = sizeof(nums) / sizeof(nums[0]);\
		head = arrayToList(nums, count);\
		printList(head);\
	    head = deleteDuplicates(head);\
		printList(head);}

struct ListNode
{
    int val;
    struct ListNode *next;
};

struct ListNode *deleteDuplicates(struct ListNode *head)
{
    struct ListNode dummy;
    struct ListNode *p, *q, *prev;
    prev = &dummy;
    dummy.next = head;
    p = q = head;
    while (p != NULL)
    {
        while (q != NULL && q->val == p->val)
		{
		    q = q->next;
		}
		if (p->next == q)
		{
		    prev = p;
		}
		else
		{
		    prev->next = q;
		}
		p = q;
    }
    return dummy.next;
}

struct ListNode *arrayToList(int* nums, int count)
{
    struct ListNode *head = NULL;
    struct ListNode *prev = NULL;
    struct ListNode *p;
    if (count == 0) return NULL;
    for (int i = 0; i < count; i++)
    {
        p = (ListNode*)malloc(sizeof(*p));
        p->val = nums[i];
        p->next = NULL;
        if (head == NULL)
        {
            head = p;
            prev = head;
        }
        else
        {
            prev->next = p;
            prev = p;
        }
    }
    return head;
}

void printList(ListNode *head)
{
	ListNode *p = head;
    while (p != NULL)
    {
        printf("%d->", p->val);
        p = p->next;
    }
    printf("null\n");
}

int main()
{
    struct ListNode *head;
    int count;
    int head1[] = {1,2,3,3,4,4,5};
	Test(head1)
    int head2[] = {1,1,1,2,3};
	Test(head2)
    return 0;
}

输出：

1->2->3->3->4->4->5->null
1->2->5->null
1->1->1->2->3->null
2->3->null 

---

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