笔记
简介
快速排序：确定分界数，左边小于分界，右边大于分界数，通过递归来不断重置分界数划分区域，直至完成排序
时间复杂度
最优 n*logn
最差 n^2
空间复杂度
原地排序，所以空间复杂度是 O(1)
解题模版
细节不在阐述，自己理解一下
class Solution {
public void sortColors(int[] nums) {
if(nums == null || nums.length <=1)
return ;
quicksort(nums,0,nums.length-1);
}
private void quicksort(int[] nums,int left,int right){
if(left>right){
return;
}
int pivot = partition(nums,left,right);
quicksort(nums,left,pivot-1);
quicksort(nums,pivot+1,right);
}
private int partition(int[] nums,int left,int right){
int pivot1 = nums[right];
int i = left-1;
for(int j = left;j<right;j++){
if(nums[j]<=pivot1)
i++;
swap(nums,i,j);
}
swap(nums,i+1,right);
return i+1;
}
private void swap(int[] nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
练习题
力扣912. 排序数组
套用模版，完美解决
class Solution {
public int[] sortArray(int[] nums) {
if(nums == null || nums.length <= 1){
return nums;
}
quickSort(nums,0,nums.length-1);
return nums;
}
private void quickSort(int[] nums,int start,int end){
if(start >= end){
return;
}
int pivot = partition(nums,start,end);
quickSort(nums,start,pivot-1);
quickSort(nums,pivot+1,end);
}
private int partition(int []nums,int start,int end){
int pivot = nums[end];
int i = start -1;
for(int j =start;j<end;j++){
if(nums[j]<= pivot){
i++;
swap(nums,i,j);
}
}
swap(nums,i+1,end);
return i+1;
}
private void swap(int[] nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
力扣215 数组中的第K个最大元素
题中要求使用O(n)时间复杂度，因此可以借助快排的分区思想，因为求第K大，并不是全部排序，平均下来时间复杂度满足条件。
class Solution {
public int findKthLargest(int[] nums, int k) {
if(nums == null || nums.length == 0 || k<=0 || k>nums.length)
return -1;
int targetIndex = nums.length-k;
int left =0;
int right = nums.length-1;
while(true){
int pivotIndex = partition(nums,left,right);
if(pivotIndex == targetIndex){
return nums[pivotIndex];
}else if(pivotIndex <targetIndex){
left = pivotIndex +1;
}else{
right = pivotIndex-1;
}
}
}
private int partition(int [] nums,int left,int right){
int pivot = nums[right];
int i = left-1;
for(int j =left;j<right;j++){
if(nums[j]<=pivot){
i++;
swap(nums,i,j);
}
}
swap(nums,i+1,right);
return i+1;
}
private void swap(int[] nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j]= temp;
}
}
看到题目要求，非常适合使用快排，直接默写
class Solution {
public void sortColors(int[] nums) {
if(nums == null || nums.length <=1)
return ;
quicksort(nums,0,nums.length-1);
}
private void quicksort(int[] nums,int left,int right){
if(left>right){
return;
}
int pivot = partition(nums,left,right);
quicksort(nums,left,pivot-1);
quicksort(nums,pivot+1,right);
}
private int partition(int[] nums,int left,int right){
int pivot1 = nums[right];
int i = left-1;
for(int j = left;j<right;j++){
if(nums[j]<=pivot1){
i++;
swap(nums,i,j);
}
}
swap(nums,i+1,right);
return i+1;
}
private void swap(int[] nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}