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数组分块如二叉树的前序遍历, 而归并排序就如二叉树的后序遍历
912. 排序数组
解法
使用归并算法
- 根据中间点划分区间, mid =(right + left ) / 2
- 将左右区间排序
- 合并两个有序数组
- 处理没有遍历完的数组
- 将排好序的数组tmp覆盖到原数组nums里
优化: 将tmp数组new成一个全局变量,减少频繁的空间开销
代码
class Solution {
int[] tmp;
public int[] sortArray(int[] nums) {
int n = nums.length;
tmp = new int[n];
mergeSort(nums, 0, n - 1);
return nums;
}
public void mergeSort(int[] nums, int left, int right){
if(left >= right) return;
//1.找到中间点mid
int mid = (left + right) / 2;
//2.左右排好序
mergeSort(nums, left, mid);
mergeSort(nums, mid + 1, right);
//3.合并有序数组 [left, mid] [mid + 1, right]
int cur1 = left, cur2 = mid + 1, i = 0;
while(cur1 <= mid && cur2 <= right){
tmp[i++] = nums[cur1] <= nums[cur2] ? nums[cur1++] : nums[cur2++];
}
//处理没有遍历完的数组
while(cur1 <= mid) tmp[i++] = nums[cur1++];
while(cur2 <= right) tmp[i++] = nums[cur2++];
//4.tmp覆盖nums
for(int j = left; j <= right; j++){
nums[j] = tmp[j - left];
}
}
}LCR 170. 交易逆序对的总数（数组中的逆序对）
解法
解法一: 暴力枚举: 两层for循环
解法二: 分治,N*logN
将数组通过mid分为两部分,
- 先找左半部分的逆序对, 再左排序
- 再找右半部分的逆序对, 再右排序
- 最后找一左一右组合的逆序对(此时左数组和右数组分别都是有序的),再合并排序
策略一: 找出该数(nums[cur1] )之前,有多少个(mid - cur2 + 1)数比我(nums[cur2] )大.
- 当nums[cur1] <= nums[cur2] 时 , cur1++;
- 当nums[cur1] > nums[cur2]时 ,ret += mid - cur1 + 1, cur2++
策略二: 找出该数(nums[cur1] )之后,有多少个(right - cur2 + 1)数比我(nums[cur2] )小.
- 当nums[cur1] <= nums[cur2] 时 , cur2++;
- 当nums[cur1] > nums[cur2]时 ,ret += right - cur2 + 1, cur1++
- 时间复杂度：O(nlogn)，这是归并排序的时间复杂度，其中 n 是数组的长度。
- 空间复杂度：O(n)，主要是临时数组
tmp的空间开销。
代码
class Solution {
int[] tmp;
public int reversePairs(int[] nums) {
int n = nums.length;
tmp = new int[n];
return merge(nums, 0, n - 1);
}
public int merge(int[] nums, int left, int right){
if(left >= right) return 0;
//找出中间点mid
int mid = (left + right) / 2;
int ret = 0;
//左右排好序,返回结果
ret += merge(nums, left, mid);
ret += merge(nums, mid + 1, right);
//合并有序数组,计算一左一右逆序对
int cur1 = left, cur2 = mid + 1, i = 0;
while(cur1 <= mid && cur2 <= right){
if(nums[cur1] <= nums[cur2]){
tmp[i++] = nums[cur1++];
}else{
ret += mid - cur1 + 1;
tmp[i++] = nums[cur2++];
}
}
//处理剩下的数组
while(cur1 <= mid) tmp[i++] = nums[cur1++];
while(cur2 <= right) tmp[i++] = nums[cur2++];
for(int j = left; j <= right; j++){
nums[j] = tmp[j - left];
}
return ret;
}
}315. 计算右侧小于当前元素的个数
解法
解法: 归并排序
和上一题类似,但是这里只能通过策略二: 降序解决
注意: 最终结果是加在原始数组的下标中
代码
class Solution {
int[] ret;
int[] index; //标记 nums中当前元素的原始下标
int[] tmpIndex;
int[] tmpNums;
public List<Integer> countSmaller(int[] nums) {
int n = nums.length;
ret = new int[n];
index = new int[n];
tmpIndex = new int[n];
tmpNums = new int[n];
//初始化index数组
for(int i = 0; i < n; i++){
index[i] = i;
}
merge(nums, 0, n - 1);
List<Integer> l = new ArrayList<>();
for(int x : ret){
l.add(x);
}
return l;
}
public void merge(int[] nums, int left, int right){
if(left >= right) return;
int mid = (left + right) / 2;
merge(nums, left, mid);
merge(nums, mid + 1, right);
//[left, mid] [mid + 1, right] 处理一左一右的情况
int cur1 = left, cur2 = mid + 1, i = 0;
while(cur1 <= mid && cur2 <= right){//降序排列
if(nums[cur1] <= nums[cur2]){
tmpNums[i] = nums[cur2];
tmpIndex[i++] = index[cur2++];//下标数组同步修改
}else{
ret[index[cur1]] += right - cur2 + 1;//结果在原数组下标对应值位置上
tmpNums[i] = nums[cur1];
tmpIndex[i++] = index[cur1++];
}
}
while(cur1 <= mid) {
tmpNums[i] = nums[cur1];
tmpIndex[i++] = index[cur1++];
}
while(cur2 <= right){
tmpNums[i] = nums[cur2];
tmpIndex[i++] = index[cur2++];
}
for(int j = left; j <= right; j++){
nums[j] = tmpNums[j - left];
index[j] = tmpIndex[j - left];
}
}
}493. 翻转对
解法
注意: 要先统计结果后,再进行排序
边界情况, 可以将2倍的数转为long,也可以使用 / 2.0
代码
策略一: 降序,谁大谁在前面
class Solution {
int[] tmp;
public int reversePairs(int[] nums) {
int n = nums.length;
tmp = new int[n];
return merge(nums, 0, n - 1);
}
public int merge(int[] nums, int left, int right){
if(left >= right) return 0;
int mid = (left + right) / 2, ret = 0;
ret += merge(nums, left, mid);
ret += merge(nums, mid + 1, right);
//[left, mid] [mid + 1, right]
//先统计,后归并
int x = left, y = mid + 1;
while(x <= mid && y <= right){//降序
if(nums[x] / 2.0 > nums[y] ){
ret += right - y + 1;
x++;
}else{
y++;
}
}
int cur1 = left, cur2 = mid + 1, i = 0;
while(cur1 <= mid && cur2 <= right){
if(nums[cur1] <= nums[cur2]){
tmp[i++] = nums[cur2++];
}else{
tmp[i++] = nums[cur1++];
}
}
while(cur1 <= mid) tmp[i++] = nums[cur1++];
while(cur2 <= right) tmp[i++] = nums[cur2++];
for(int j = left; j <= right; j++){
nums[j] = tmp[j - left];
}
return ret;
}
}策略二: 升序
class Solution {
int[] tmp;
public int reversePairs(int[] nums) {
int n = nums.length;
tmp = new int[n];
return merge(nums, 0, n - 1);
}
public int merge(int[] nums, int left, int right){
if(left >= right) return 0;
int mid = (left + right) / 2, ret = 0;
ret += merge(nums, left, mid);
ret += merge(nums, mid + 1, right);
//[left, mid] [mid + 1, right]
//先统计,后归并
int x = left, y = mid + 1;
while(x <= mid && y <= right){//降序
if(nums[x] / 2.0 > nums[y] ){
ret += mid - x + 1;
y++;
}else{
x++;
}
}
int cur1 = left, cur2 = mid + 1, i = 0;
while(cur1 <= mid && cur2 <= right){
if(nums[cur1] <= nums[cur2]){
tmp[i++] = nums[cur1++];
}else{
tmp[i++] = nums[cur2++];
}
}
while(cur1 <= mid) tmp[i++] = nums[cur1++];
while(cur2 <= right) tmp[i++] = nums[cur2++];
for(int j = left; j <= right; j++){
nums[j] = tmp[j - left];
}
return ret;
}
}
