一刷
Version #1 Quick sort
这个写的不好，看下边一个版本
有两个问题
1.不应该选取nums[start] 作为pivot，因为如果原有的nums是descending or ascending order, the time efficiency will be reduced
2.left++和right--的条件应该是大于或者小于
Runtime: 952 ms, faster than 16.06% of Java online submissions for Sort an Array.
Memory Usage: 72.9 MB, less than 17.00% of Java online submissions for Sort an Array.
Version #1 Quick Sort
class Solution {
public int[] sortArray(int[] nums) {
// Quick sort
// Select the 1st element in the array, devide the array by comparing each element with the seleted element
// start end
// nums[i] <= p nums[i] >= p
return sortArrayHelper(nums, 0, nums.length - 1);
}
// s e r l
// [0,1,1,1,1,1,1,1,1,6,5]
private int[] sortArrayHelper(int[] nums, int start, int end) {
if (start >= end) {
return nums;
}
// lr
// [(1),1,1,1,1,1,1,1,0,6,5]
int left = start + 1, right = end;
int p = nums[start];
while (left <= right) {
// left is the first character that's larger than p
while (left <= right && nums[left] <= p) {
left++;
}
while (left <= right && nums[right] >= p) {
right--;
}
if (left <= right) {
swap(nums, left, right);
left++;
right--;
}
}
swap(nums, start, left - 1);
sortArrayHelper(nums, start, left - 2);
sortArrayHelper(nums, left, end);
return nums;
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
Version #1 Quick sort
Optimized version
Using nums[(start + end) / 2] as the pivot
This will improve the speed if the nums are in ascending or descending order
Runtime: 10 ms, faster than 88.81% of Java online submissions for Sort an Array.
Memory Usage: 50.8 MB, less than 97.00% of Java online submissions for Sort an Array.
class Solution {
public int[] sortArray(int[] nums) {
// Quick sort
// Shuffle the array (optional)
// Partition so that, for some j
// - entry a[j] is in place
// - no larger entry to the left of j
// - no smaller entry to the right of j
// Sort each piece recursively
if (nums == null) {
return null;
}
quickSort(nums, 0, nums.length - 1);
return nums;
}
// Sort given array from index start to index end
private void quickSort(int[] nums, int start, int end) {
if (start >= end) {
return;
}
int pivot = nums[start + (end - start) / 2];
// partition the subarray between start and end
// so that all nums <= pivot are on the left, all nums >= pivot are on the right
int left = start, right = end;
while (left <= right) {
while (left <= right && nums[left] < pivot) {
left++;
}
while (left <= right && nums[right] > pivot) {
right--;
}
if (left <= right) {
// swap nums[left] and nums[right]
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
left++;
right--;
}
}
// All nums to the left side of pointer l are smaller than pivot
// All nums to the right side of pointer r are larger than pivot
// l
// r
// 0 1 2 3
quickSort(nums, start, right);
quickSort(nums, left, end);
}
}
Version #2 Merge Sort
每次都new了一个新的result数组，浪费了空间，同时time wasted during the construction of new temporary arrays
下面写了optimized solution
Runtime: 19 ms, faster than 52.47% of Java online submissions for Sort an Array.
Memory Usage: 72.8 MB, less than 17.01% of Java online submissions for Sort an Array.
class Solution {
public int[] sortArray(int[] nums) {
// Merge sort
// Divide nums array from the middle, sort each subarray separately
// Merge two arrays together
if (nums == null) {
return null;
}
return mergeSort(nums, 0, nums.length - 1);
}
private int[] mergeSort(int[] nums, int start, int end) {
// exit criteria
if (start > end) {
return new int[0];
}
if (start == end) {
return new int[]{nums[start]};
}
int[] leftNums = mergeSort(nums, start, (start + end) / 2);
int[] rightNums = mergeSort(nums, (start + end) / 2 + 1, end);
return merge(leftNums, rightNums);
}
private int[] merge(int[] nums1, int[] nums2) {
int[] result = new int[nums1.length + nums2.length];
int p1 = 0, p2 = 0, pr = 0;
while (p1 < nums1.length && p2 < nums2.length) {
if (nums1[p1] <= nums2[p2]) {
result[pr] = nums1[p1];
p1++;
pr++;
} else {
result[pr] = nums2[p2];
p2++;
pr++;
}
}
while (p1 < nums1.length) {
result[pr] = nums1[p1];
pr++;
p1++;
}
while (p2 < nums2.length) {
result[pr] = nums2[p2];
pr++;
p2++;
}
return result;
}
}
Version #2 Merge Sort [Optimized]
To preserve stability, if two keys are equal, we always need to first choose from the left subarray
Runtime: 10 ms, faster than 88.81% of Java online submissions for Sort an Array.
Memory Usage: 51.2 MB, less than 91.77% of Java online submissions for Sort an Array.
class Solution {
public int[] sortArray(int[] nums) {
// Merge sort
// Divide array into two halves
// Recursively sort each half
// Merge two halves
if (nums == null) {
return null;
}
// Create a auxiliary array to temporarily store the subarray
int[] aux = new int[nums.length];
quickSort(nums, 0, nums.length - 1, aux);
return nums;
}
// quickSort - sort array from index start to index end
private void quickSort(int[] nums, int start, int end, int[] aux) {
// base case
if (start >= end) { // stop when there's less than or equal to one element
return;
}
int mid = start + (end - start) / 2;
quickSort(nums, start, mid, aux);
quickSort(nums, mid + 1, end, aux);
// Optimize - if the largest element in the first half is smaller than the smallest element in the second half, no need to merge since the array is already sorted
if (nums[mid] < nums[mid + 1]) {
return;
}
merge(nums, start, mid, end, aux);
}
// merge - merge subarray [start, mid] with subarray [mid + 1, end]
// Guarantee that the two subarrays are ordered
private void merge(int[]nums, int start, int mid, int end, int[] aux) {
// Firstly copy the original array to the auxiliary array
for (int i = start; i <= end; i++) {
aux[i] = nums[i];
}
// keep 3 pointers, pFirstHalf, pSecondHalf, pResult
int pFirstHalf = start, pSecondHalf = mid + 1, pResult = start;
while (pFirstHalf <= mid && pSecondHalf <= end) {
// If two elements are equal, choose from the firsthalf
if (aux[pFirstHalf] <= aux[pSecondHalf]) {
nums[pResult] = aux[pFirstHalf];
pFirstHalf++;
} else {
nums[pResult] = aux[pSecondHalf];
pSecondHalf++;
}
pResult++;
}
while (pFirstHalf <= mid) {
nums[pResult++] = aux[pFirstHalf++];
}
while (pSecondHalf <= end) {
nums[pResult++] = aux[pSecondHalf++];
}
}
}
