912.排序数组
快速排序
思路：
1. 设置一个pivot2. 将小于nums[pivot]的值 放在左边3. 将 大于nums[pivot]的值 放在 右边4. 递归调用
注意：必须先比较nums[high] 与pivot
代码：
class Solution {int partition(vector<int>&nums, int low, int high){int pivot = nums[low];while(low<high){while(low <high && nums[high]>=pivot){high--;}nums[low] = nums[high];while(low<high && nums[low]<=pivot){low++;}nums[high] = nums[low];}nums[low] = pivot;return low;}void quiksort(vector<int>&nums, int low, int high){if(low<high){int pos = partition(nums, low, high);quiksort(nums, low, pos-1);quiksort(nums, pos+1, high);}}public:vector<int> sortArray(vector<int>& nums) {quiksort(nums, 0, (int)nums.size()-1);return nums;}
};
堆排序
思路：
1.构造大顶堆
2. 取出堆顶元素
3. 构造大顶堆
4. 取出堆顶元素 将堆顶元素放入末尾，下次构造大顶堆时不考虑
注意：
代码：
class Solution {void buildMaxHeap(vector<int>& nums) {int n = nums.size();for (int i = (n - 1) / 2; i >= 0; --i) {maxHeapify(nums, i, n);}}void maxHeapify(vector<int>& nums, int i, int n) {while (i * 2 + 1 < n) {// 代表当前 i 节点的左右儿子；// 超出数组大小则代表当前 i 节点为叶子节点，不需要移位int lSon = 2 * i + 1;int rSon = 2 * i + 2;int large = i;if (lSon < n && nums[lSon] > nums[i]) large = lSon;if (rSon < n && nums[rSon] > nums[large]) large = rSon;if (large != i) {swap(nums[i], nums[large]);// 迭代判断对应子节点及其儿子节点的大小关系i = large;} else {break;}}}public:vector<int> sortArray(vector<int>& nums) {// heapSort 堆排序int n = nums.size();// 将数组整理成大根堆buildMaxHeap(nums);for (int i = n - 1; i >= 1; --i) {// 依次弹出最大元素，放到数组最后，当前排序对应数组大小 - 1swap(nums[0], nums[i]);--n;maxHeapify(nums, 0, n);}return nums;}
};归并排序
思路： 将有序序列 合并成一个
注意：
代码：
class Solution {vector<int> tmp;void mergeSort(vector<int>& nums, int l, int r) {if (l >= r) return;int mid = (l + r) >> 1;mergeSort(nums, l, mid);mergeSort(nums, mid + 1, r);int i = l, j = mid + 1;int cnt = 0;while (i <= mid && j <= r) {if (nums[i] <= nums[j]) {tmp[cnt++] = nums[i++];}else {tmp[cnt++] = nums[j++];}}while (i <= mid) {tmp[cnt++] = nums[i++];}while (j <= r) {tmp[cnt++] = nums[j++];}for (int i = 0; i < r - l + 1; ++i) {nums[i + l] = tmp[i];}}
public:vector<int> sortArray(vector<int>& nums) {tmp.resize((int)nums.size(), 0);mergeSort(nums, 0, (int)nums.size() - 1);return nums;}
};
)
