1、vector的使用
#define _CRT_SECURE_NO_WARNINGS 1
#include <iostream>
#include <vector>
using namespace std;
void Test1()
{
vector<int> v1;
vector<int> v2(10, 15);
vector<int> v3(v2.begin(), v2.end());
string str("hello world");
vector<char> v4(str.begin(), str.end());
vector<char> v5(v4);
//[下标]
for (size_t i = 0; i < v3.size(); i++)
{
cout << v3[i] << ' ';
}
cout << endl;
//迭代器
//vector<char>::iterator it = v4.begin();
auto it = v4.begin();
while (it != v4.end())
{
cout << *it;
it++;
}
cout << endl;
//范围for
for (auto ch : v5)
{
cout << ch;
}
}
void Test2()
{
vector<int> v;
//v.reserve(100); //reserve，size没变，还是从0开始
size_t sz = v.capacity(); //初始容量
for (int i = 0; i < 100; i++)
{
v.push_back(i);
if (sz != v.capacity())
{
sz = v.capacity();
cout << "扩容：" << sz << endl;
}
}
}
void Test3()
{
vector<int> v;
cout << v.max_size() << endl;
//v.reserve(100); //error【size = 0 capacity 100】
v.resize(100); //【size = 100 capacity 100】【只能操控size以内的数据进行赋值、修改等】
for (size_t i = 0; i < 100; i++)
{
v[i] = i;
}
for (auto e : v)
{
cout << e << " ";
}
cout << endl;
}
void Test4()
{
vector<int> v1;
v1.push_back(1);
v1.push_back(2);
v1.push_back(3);
v1.push_back(4);
for (auto e : v1)
{
cout << e << ' ';
}
cout << endl;
//插入
v1.insert(v1.begin(), 5); //提供的是地址，不是下标，在该位置前插入
for (auto e : v1)
{
cout << e << ' ';
}
cout << endl;
//vector没有提供find，直接用算法库里的
auto it = find(v1.begin(), v1.end(), 3);
if (it != v1.end()) //【迭代器传区间都是左闭右开】
{
v1.insert(it, 30);
}
for (auto e : v1)
{
cout << e << ' ';
}
cout << endl;
//删除
it = find(v1.begin(), v1.end(), 3);
if (it != v1.end())
{
v1.erase(it);
}
for (auto e : v1)
{
cout << e << ' ';
}
cout << endl;
}
void Test5()
{
vector<int> v;
v.push_back(1);
v.push_back(2);
v.push_back(3);
v.push_back(4);
cout << "v.size() = " << v.size() << endl;
cout << "v.capacity() = " << v.capacity() << endl;
v.clear(); //只清数据（size），不清capacity
cout << "clear后size = " << v.size() << endl;
cout << "clear后capacity = " << v.capacity() << endl;
}
int main()
{
Test5();
return 0;
}
2、vector的OJ题
（1）二叉树的前序遍历
class Solution
{
public:
void preorder(TreeNode* root, vector<int>& v)
{
if (root == nullptr)
{
return;
}
v.push_back(root->val);
preorder(root->left, v);
preorder(root->right, v);
}
vector<int> preorderTraversal(TreeNode* root)
{
vector<int> v1;
preorder(root, v1);
return v1;
}
};
（2）只出现一次的数字Ⅰ
class Solution
{
public:
int singleNumber(vector<int>& nums)
{
size_t val = 0;
for (auto e : nums)
{
val ^= e;
}
return val;
}
};
（3）杨辉三角
class Solution
{
public:
vector<vector<int>> generate(int numRows)
{
vector<vector<int>> vv;
vv.resize(numRows);
for (size_t i = 0; i < vv.size(); i++)
{
vv[i].resize(i + 1);
vv[i][0] = vv[i][vv[i].size() - 1] = 1;
}
for (size_t i = 0; i < numRows; i++)
{
for (size_t j = 0; j < vv[i].size(); j++)
{
if (vv[i][j] != 1)
{
vv[i][j] = vv[i - 1][j] + vv[i - 1][j - 1];
}
}
}
return vv;
}
};
（4）电话号码的字母组合
class Solution
{
const char* numStrArr[10] = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };
public:
void combine(string& digits, int i, string combine_str, vector<string>& ret)
{
if (i == digits.size())
{
ret.push_back(combine_str);
return;
}
int num = digits[i] - '0'; //每次从digits取一个字符转成数字
string str = numStrArr[num]; //记录该数字对应的字符串
/*for (int j = 0; j < str.size(); j++)
{
combine(digits, i + 1, combine_str + str[j], ret);
}*/
for (auto ch : str)
{
combine(digits, i + 1, combine_str + ch, ret);
}
}
vector<string> letterCombinations(string digits)
{
vector<string> v; //用来存放所有组合结果
if (digits.empty())
{
return v;
}
string str; //创建一个空字符串用来存每次的组合结果
combine(digits, 0, str, v);
return v;
}
};
（5）删除有序数组中的重复项
class Solution
{
public:
//方法1：双指针+vector的使用
/*int removeDuplicates(vector<int>& nums)
{
int slow = 0;
int fast = 1;
while (fast < nums.size())
{
if (nums[slow] == nums[fast])
{
nums.erase(nums.begin() + fast);
}
else
{
fast++;
slow++;
}
}
return nums.size(); //返回元素个数
}*/
//方法2：双指针
int removeDuplicates(vector<int>& nums)
{
int slow = 0;
int fast = 1;
while (fast < nums.size())
{
if (nums[fast] == nums[slow])
{
fast++;
}
else
{
slow++;
nums[slow] = nums[fast];
fast++;
}
}
return slow + 1; //返回元素个数=下标+1
}
};
（6）只出现一次的数字Ⅱ
class Solution
{
public:
void Swap(int* a, int* b)
{
int tmp = *a;
*a = *b;
*b = tmp;
}
int Partsort(int* a, int left, int right)
{
int keyi = left;
while (left < right)
{
//右边先开始
while (left < right && a[right] >= a[keyi])
{
right--;
}
//右边找到再找左边
while (left < right && a[left] <= a[keyi])
{
left++;
}
Swap(&a[left], &a[right]);
}
Swap(&a[left], &a[keyi]);
return left;
}
void nums_Sort(int* a, int begin, int end)
{
if (begin >= end)
{
return;
}
int keyi = Partsort(a, begin, end);
nums_Sort(a, begin, keyi - 1);
nums_Sort(a, keyi + 1, end);
}
//方法1：【先排序，再3个3个遍历数组】
int singleNumber(vector<int>& nums)
{
//快排
nums_Sort(&nums[0], 0, nums.size() - 1);
int result;
for (int i = 0; i < nums.size(); i += 3)
{
//判出界
if (i + 1 >= nums.size())
{
result = nums[i];
break;
}
if (nums[i] != nums[i + 2])
{
result = nums[i];
break;
}
}
return result;
}
//方法2：【按位计算】
/*int singleNumber(vector<int>& nums)
{
int result = 0;
for (int i = 0; i < 32; i++)
{
int sum = 0;
for (int num : nums)
{
sum += ((num >> i) & 1);
}
if (sum % 3)
{
result |= (1 << i); //【|=按位或】
}
}
return result;
}*/
};
（7）只出现一次的数字Ⅲ
博客：👉【C/C++】x & -x 的含义
防止溢出：
- INT_MAX = 2^31-1 = 2147483647
- INT_MIN = -2^31 = -2147483648
class Solution
{
public:
vector<int> singleNumber(vector<int>& nums)
{
//算出所有数异或后的和
int xorsum = 0;
for (auto e : nums)
{
xorsum ^= e;
}
//防止溢出
//测试用例：[1,1,0,-2147483648]，如果对-2147483648取负，那就溢出了
int lsb = (xorsum == INT_MIN ? xorsum : xorsum & (-xorsum)); //取出xorsum的二进制表示中最低位那个1
//按L位的不同分成两个组
int type1 = 0, type2 = 0;
for (auto f : nums)
{
if (f & lsb)
{
type1 ^= f; //二进制表示的第L位为1的数
}
else
{
type2 ^= f; //进制表示的第L位为0的数
}
}
return { type1,type2 };
}
};
（8）数组中出现次数超过一半的数字
class Solution
{
public:
int MoreThanHalfNum_Solution(vector<int>& numbers)
{
sort(numbers.begin(),numbers.end()); //排序数组
return numbers[numbers.size()/2]; //取中间元素
}
};
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