【C++】vector的使用
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1、vector的使用
#define _CRT_SECURE_NO_WARNINGS 1#include ::iterator it = v4.begin(); auto it = v4.begin();while (it != v4.end()){cout << *it;it++;}cout << endl;//范围forfor (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capacitycout << "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 (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 //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】 };（7）只出现一次的数字Ⅲ
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 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 }};来源地址：https://blog.csdn.net/weixin_69231613/article/details/132229003
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