1. 题目

给你一个整数数组 nums，将该数组升序排列。

示例 1：
输入：nums = [5,2,3,1]
输出：[1,2,3,5]示例 2：
输入：nums = [5,1,1,2,0,0]
输出：[0,0,1,1,2,5]提示：
1 <= nums.length <= 50000
-50000 <= nums[i] <= 50000

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sort-an-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

基础的排序算法，写一遍复习一下。
参考我的博客：
10种C++排序算法
快速排序quicksort算法优化
快速排序quicksort算法细节优化（一次申请内存/无额外内存排序）

2.1 插入排序

class Solution {
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;int i, j;for(i = 0; i < arr.size(); ++i){for(j = i; j > 0; --j){if(arr[j-1] > arr[j])swap(arr[j-1],arr[j]);elsebreak;}}return arr;}
};

9 / 10 个通过测试用例，超时

2.2 冒泡排序

class Solution {
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;int i, j;bool arrSorted = false;for(i = 0; i < arr.size(); ++i){arrSorted = true;for(j = 1; j <= arr.size()-1-i; ++j){if(arr[j-1] > arr[j]){swap(arr[j-1],arr[j]);arrSorted = false;}}if(arrSorted)break;}return arr;}
};

超时

2.3 选择排序

class Solution {	//选择
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;int i, j, minIdx = 0;for(i = 0; i < arr.size()-1; ++i){minIdx = i;for(j = i+1; j < arr.size(); ++j){if(arr[minIdx] > arr[j])minIdx = j;}swap(arr[i],arr[minIdx]);}return arr;}
};

超时

2.4 希尔排序

class Solution {	//希尔
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;int i, j, gap = 1;for(gap = arr.size()/2; gap > 0; gap /= 2){for(i = gap; i < arr.size(); ++i){for(j = i; j-gap >= 0 && arr[j-gap]>arr[j]; j -= gap)swap(arr[j-gap],arr[j]);}}return arr;}
};

72 ms 15.8 MB

2.5 归并排序

class Solution {	//归并vector<int> temp;
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;temp.resize(arr.size());mergeSort(arr,0, arr.size()-1);return arr;}void mergeSort(vector<int>& arr, int l, int r){if(l == r)return;int mid = l+((r-l)>>1);mergeSort(arr,l,mid);mergeSort(arr,mid+1,r);merge(arr,l,mid,r);}void merge(vector<int>& arr, int l, int mid, int r){int i = l, j = mid+1, k = 0;while(i <= mid && j <= r){if(arr[i] <= arr[j])temp[k++] = arr[i++];elsetemp[k++] = arr[j++];}while(i <= mid)temp[k++] = arr[i++];while(j <= r)temp[k++] = arr[j++];for(k = 0, i = l; i <= r; ++i,++k)arr[i] = temp[k];}
};

52 ms 16 MB

2.6 快速排序

class Solution {	//快排
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;qsort(arr,0, arr.size()-1);return arr;}void qsort(vector<int>& arr, int l, int r){if(l >= r)return;int Pl = l, Pr = l;partition(arr,l,r,Pl,Pr);qsort(arr,l,Pl-1);qsort(arr,Pr+1,r);}void partition(vector<int>& arr, int l, int r, int& Pl, int& Pr){	selectMid(arr,l,r);int P = arr[l];int i = l, j = r;while(i < j){while(i < j && P < arr[j])//没有等于号，哨兵都在左侧j--;swap(arr[i], arr[j]);while(i < j && arr[i] <= P)i++;swap(arr[i], arr[j]);}Pl = Pr = i;for(i = i-1; i >= l; --i)//聚集左侧与哨兵相等的到中间{if(arr[i] == P){Pl--;swap(arr[i], arr[Pl]);}}}void selectMid(vector<int>& arr, int l, int r){int mid = l+((r-l)>>1);if(arr[mid] > arr[r])swap(arr[mid],arr[r]);if(arr[l] > arr[r])swap(arr[l], arr[r]);if(arr[mid] > arr[l])swap(arr[mid], arr[l]);}
};

52 ms 15.6 MB
96 ms 15.8 MB（删除聚集哨兵操作后的用时）

2.7 堆排序

class Solution {	//堆排序, 建堆（升序建大堆，降序建小堆）
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;for(int i = arr.size()/2-1; i >= 0; --i)adjust(arr,i,arr.size());//建堆for(int i = arr.size()-1; i >= 0; --i){swap(arr[i],arr[0]);adjust(arr,0,i);}return arr;}void adjust(vector<int>& arr, int i, int len){int lchild = 2*i+1, rchild = 2*i+2, maxIdx = i;while(lchild < len){if(lchild < len && arr[lchild] > arr[maxIdx])maxIdx = lchild;if(rchild < len && arr[rchild] > arr[maxIdx])maxIdx = rchild;if(maxIdx != i){swap(arr[i],arr[maxIdx]);lchild = 2*maxIdx+1;rchild = lchild+1;i = maxIdx;}elsebreak;}}
};

72 ms 15.8 MB

2.8 计数排序

class Solution {	//计数排序
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;int i, j = 0, min, max;min = max = arr[0];for(i = 1; i < arr.size(); ++i){min = arr[i] < min ? arr[i] : min;max = arr[i] > max ? arr[i] : max;}const int N = max-min+1;vector<int> count(N,0);for(i = 0; i < arr.size(); ++i)count[arr[i]-min]++;for(i = 0; i < N; ++i){while(count[i]--)arr[j++] = i+min;}return arr;}
};

36 ms 16.1 MB

2.9 桶排序

• 数据装桶，桶内快排

class Solution {	//桶排序
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;int i, j = 0, min, max;min = max = arr[0];for(i = 1; i < arr.size(); ++i){min = arr[i] < min ? arr[i] : min;max = arr[i] > max ? arr[i] : max;}if(min == max)return arr;int div = 1000;//桶个数int space = (max-min)/div+1;vector<int> temp(arr.size());vector<int> bucketsize(div,0);vector<int> bucketPos(div,0);for(i = 0; i < arr.size(); ++i)bucketsize[(arr[i]-min)/space]++;bucketPos[0] = bucketsize[0];for(i = 1; i < div; ++i)bucketPos[i] += bucketPos[i-1] + bucketsize[i];//桶结束位置的下一个for(i = 0; i < arr.size(); ++i)temp[--bucketPos[(arr[i]-min)/space]] = arr[i];for(i = 0; i < div; ++i){if(bucketsize[i] > 1){qsort(temp,bucketPos[i],bucketPos[i]+bucketsize[i]-1);}}for(i = 0; i < arr.size(); ++i)arr[i] = temp[i];return arr;}void qsort(vector<int>& arr, int l, int r){if(l >= r)return;int Pl = l, Pr = l;partition(arr,l,r,Pl,Pr);qsort(arr,l,Pl-1);qsort(arr,Pr+1,r);}void partition(vector<int>& arr, int l, int r, int& Pl, int& Pr){selectMid(arr,l,r);int P = arr[l];int i = l, j = r;while(i < j){while(i < j && P < arr[j])//没有等于号，哨兵都在左侧j--;swap(arr[i], arr[j]);while(i < j && arr[i] <= P)i++;swap(arr[i], arr[j]);}Pl = Pr = i;for(i = i-1; i >= l; --i){if(arr[i] == P){Pl--;swap(arr[i], arr[Pl]);}}}void selectMid(vector<int>& arr, int l, int r){int mid = l+((r-l)>>1);if(arr[mid] > arr[r])swap(arr[mid],arr[r]);if(arr[l] > arr[r])swap(arr[l], arr[r]);if(arr[mid] > arr[l])swap(arr[mid], arr[l]);}
};

40 ms 16.3 MB

2.10 基数排序

• 注意处理负数，被坑了，负数%后越界，window不报错尽然。。

class Solution {	//基数排序vector<int> temp;
public:vector<int> sortArray(vector<int>& arr) {if(arr.size() <= 1)return arr;int i, max = INT_MIN;long exp;temp.resize(arr.size());for(i = 0; i < arr.size(); ++i){arr[i] += 50000;//便于负数处理max = arr[i] > max ? arr[i] : max;}for(exp = 1; max/exp > 0; exp*=10)radix_sort(arr,exp);for(i = 0; i < arr.size(); ++i)arr[i] -= 50000;//还原return arr;}void radix_sort(vector<int>& arr, long exp){vector<int> bucketsize(10,0);int i;for(i = 0; i < arr.size(); ++i)bucketsize[(arr[i]/exp)%10]++;for(i = 1; i < 10; ++i)bucketsize[i] += bucketsize[i-1];//桶最后一个位置+1for(i = arr.size()-1; i >= 0; --i)temp[--bucketsize[(arr[i]/exp)%10]] = arr[i];for(i = 0; i < arr.size(); ++i)arr[i] = temp[i];}
};

56 ms 16 MB

3. 复杂度表

图片参考 https://leetcode-cn.com/problems/sort-an-array/solution/shi-er-chong-pai-xu-suan-fa-bao-ni-man-yi-dai-gift/

r 为排序数字的范围，d 是数字总位数，k 是数字总个数