https://www.acwing.com/solution/content/24716/
https://www.acwing.com/blog/content/3585/

在待排序序列 nums 中选取一个元素 pivot 作为枢轴,通过一趟排序将 nums 0 ~ n-1分为两部分 0 ~ k-1 k+1 ~ n-1 (使得左侧的元素 ≤ 枢轴,右侧的元素 ≥ 枢轴)。递归地对两个子部分进行上述过程,直到每部分只有一个元素或为空。
void quick_sort(int q[], int l, int r)
{
if (l >= r) return;
int i = l - 1, j = r + 1, x = q[l + r >> 1];
while (i < j)
{
do i ++ ; while (q[i] < x);
do j -- ; while (q[j] > x);
if (i < j) swap(q[i], q[j]);
}
quick_sort(q, l, j), quick_sort(q, j + 1, r);
}
void merge_sort(vector<int>& nums, int l, int r) {
if (l >= r) return;
vector<int> temp(r - l + 1);
int mid = (l + r) / 2;
merge_sort(nums, l, mid);
merge_sort(nums, mid + 1, r);
int i = l, j = mid + 1, k = 0;
while (i <= mid && j <= r) {
if (nums[i] < nums[j]) {
temp[k++] = nums[i++];
} else {
temp[k++] = nums[j++];
}
}
while (i <= mid) temp[k++] = nums[i++];
while (j <= r) temp[k++] = nums[j++];
for (i = l, j = 0; i <= r; i++, j++) nums[i] = temp[j];
}
把无序数组构建成二叉堆 循环删除堆顶元素,移到集合尾部,调节堆产生新的堆顶
完全二叉树的顺序存储
// nums[0 ... n - 1]
auto left = 2 * i + 1;
auto right = 2 * i - 1;
auto parent = (i + 1) / 2;

void head_adjust(vector<int>& nums, int k, int n) {
for (int i = 2 * k + 1; i < n; i = 2 * i + 1) {
if (i + 1 < n && nums[i] < nums[i + 1]) {
i++;
}
if (nums[k] > nums[i]) break;
else {
swap(nums[k], nums[i]);
k = i;
}
}
}
void build_max_heap(vector<int>& nums, int n) {
for (int i = (n - 1) / 2; i >= 0; i--) {
head_adjust(nums, i, n);
}
}
void head_sort(vector<int>& nums) {
int n = nums.size();
if (n <= 1) return;
build_max_heap(nums, n);
for (int i = n - 1; i > 0; i--) {
swap(nums[0], nums[i]);
head_adjust(nums, 0, i);
}
}
class Solution {
public:
vector<int> sortArray(vector<int>& nums) {
int n =nums.size();
for (int i = 0; i < n - 1; i++) {
for (int j = 0; j + 1 < n - i; j++) {
if (nums[j] > nums[j + 1]) {
swap(nums[j], nums[j + 1]);
}
}
}
return nums;
}
};
class Solution {
public:
vector<int> sortArray(vector<int>& nums) {
int n =nums.size();
for (int i = 0; i < n - 1; i++) {
bool ischanged = false;
for (int j = 0; j + 1 < n - i; j++) {
if (nums[j] > nums[j + 1]) {
swap(nums[j], nums[j + 1]);
ischanged = true;
}
}
if (!ischanged) {
cout << "break" << endl;
break;
}
}
return nums;
}
};
最好情况是原数组有序,最快情况是逆序