力扣labuladong——一刷day35
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文章目录
- 前言
- 一、力扣912. 排序数组
- 二、力扣 315. 计算右侧小于当前元素的个数
- 三、力扣493. 翻转对
- 四、力扣327. 区间和的个数
前言
就这么说吧，所有递归的算法，你甭管它是干什么的，本质上都是在遍历一棵（递归）树，然后在节点（前中后序位置）上执行代码，你要写递归算法，本质上就是要告诉每个节点需要做什么
一、力扣912. 排序数组
class Solution {
public int[] sortArray(int[] nums) {
Merge.sort(nums);
return nums;
}
}
class Merge{
public static int[] temp;
public static void sort(int[] nums){
temp = new int[nums.length];
sort(nums, 0, nums.length-1);
}
public static void sort(int[] nums, int low, int high){
if(low == high){
return;
}
int mid = low + (high-low)/2;
sort(nums, low, mid);
sort(nums,mid+1,high);
merge(nums,low,high,mid);
}
public static void merge(int[] nums, int low, int high, int mid){
for(int a = low; a <= high; a ++){
temp[a] = nums[a];
}
int i = low, j = mid+1, k = low;
for(; i <= mid && j <= high;){
if(temp[i] < temp[j]){
nums[k++] = temp[i++];
}else{
nums[k++] = temp[j++];
}
}
while(i <= mid){
nums[k++] = temp[i++];
}
while(j <= high){
nums[k++] = temp[j++];
}
}
}
二、力扣 315. 计算右侧小于当前元素的个数
class Solution {
private class Pair{
int id;
int val;
public Pair(int id, int val){
this.id = id;
this.val = val;
}
}
Pair[] temp;
int[] count;
public List<Integer> countSmaller(int[] nums) {
int n = nums.length;
temp = new Pair[n];
Pair[] arr = new Pair[n];
for(int i = 0; i < n; i ++){
arr[i] = new Pair(i,nums[i]);
}
count = new int[n];
sort(arr,0,n-1);
List<Integer> res = new ArrayList<>();
for(int i = 0; i < n; i ++){
res.add(count[i]);
}
return res;
}
public void sort(Pair[] arr, int low, int high){
if(low == high){
return;
}
int mid = low + (high-low)/2;
sort(arr,low,mid);
sort(arr,mid+1,high);
merge(arr,low,high,mid);
}
public void merge(Pair[] arr, int low, int high, int mid){
for(int i = low; i <= high; i ++){
temp[i] = arr[i];
}
for(int i = low, j = mid+1, k = low; k <= high; k ++){
if(i == mid+1){
arr[k] = temp[j++];
}else if(j == high+1){
count[temp[i].id] += j-mid-1;
arr[k] = temp[i++];
}else if(temp[i].val > temp[j].val){
arr[k] = temp[j++];
}else{
count[temp[i].id] += j-mid-1;
arr[k] = temp[i++];
}
}
}
}
三、力扣493. 翻转对
class Solution {
int[] temp;
int count = 0;
public int reversePairs(int[] nums) {
temp = new int[nums.length];
sort(nums,0, nums.length-1);
return count;
}
public void sort(int[] nums, int low, int high){
if(low == high){
return;
}
int mid = low + (high-low)/2;
sort(nums,low,mid);
sort(nums,mid+1,high);
merge(nums,low,high,mid);
}
public void merge(int[] nums, int low, int high, int mid){
for(int i = low; i <= high; i ++){
temp[i] = nums[i];
}
int end = mid+1;
for(int i = low; i <= mid; i ++){
while(end <= high && (long)nums[i] > (long)2*nums[end]){
end ++;
}
count += end - mid -1;
}
for(int i = low, j = mid+1, k = low; k <= high; k ++){
if(i == mid+1){
nums[k] = temp[j++];
}else if(j == high+1){
nums[k] = temp[i++];
}else if(temp[i] > temp[j]){
nums[k] = temp[j++];
}else{
nums[k] = temp[i++];
}
}
}
}
四、力扣327. 区间和的个数
class Solution {
int lower,upper,count=0;
long[] temp;
public int countRangeSum(int[] nums, int lower, int upper) {
this.lower= lower;
this.upper = upper;
long[] preSum = new long[nums.length+1];
for (int i = 0; i < nums.length; i++) {
preSum[i + 1] = (long)nums[i] + preSum[i];
}
temp = new long[preSum.length];
sort(preSum, 0, preSum.length-1);
return count;
}
public void sort(long[] nums, int low, int high){
if(low == high){
return;
}
int mid = low + (high-low)/2;
sort(nums,low,mid);
sort(nums,mid+1,high);
merge(nums,low,high,mid);
}
public void merge(long[] nums,int low, int high, int mid){
for(int i = low; i <= high; i ++){
temp[i] = nums[i];
}
int start = mid+1, end = mid+1;
for(int i = low; i <= mid; i ++){
while(start <= high && nums[start]-nums[i] < lower){
start ++;
}
while(end <= high && nums[end] - nums[i] <= upper){
end ++;
}
count += end-start;
}
for(int i = low, j = mid+1, k = low; k <= high; k ++){
if(i == mid+1){
nums[k] = temp[j++];
}else if(j == high+1){
nums[k] = temp[i++];
}else if(temp[i] > temp[j]){
nums[k] = temp[j++];
}else{
nums[k] = temp[i++];
}
}
}
}