思考
一定要注意归并排序的含义,思考归并的意义。
主要分为两个步骤:
- 拆分
- 每次对半分
(mid = l +r >> 1)
- 输入:
raw整块
,输出:raw左块
+raw右块
- 每次对半分
- 合并
- 每次都要对
raw左块
、raw右块
按照某种规则进行合并 - 输入:
raw左块
+raw右块
,输出:raw整块
- 每次都要对
知道两个步骤之后,可以总结其他的特点:
- 拆分阶段和合并阶段是一一对应的,只不过拆分阶段是
raw
的,合并阶段符合一定的性质(对于归并排序则满足有序性)。 - 拆分时,段内是无序的,合并时,每一段都是有序的(数值有序性)。合并是针对两个有序的段进行合并,所以会经常用到双指针算法。
- 如下图所示,在合并过程中,段内是数值有序,但是相对顺序被破坏了,而两个段之间的相对顺序是不变的。
6、7、8
相对于1、2、3
的顺序是不变的,6、7、8依然在1、2、3的左边。
几道题做下来,感觉归并排序类型题的难点在于
- 题意的转化:重点就要题意是否支持将原模型分成两半来考虑,即计算左段相对后段的某种性质。
- 合并阶段对结果的计算,比如说求逆序对,那么合并的时候如何求逆序对的个数,双重循环遍历?双指针?等等。。。
普通模板
int* merge(int l, int r) {
if (l > r) return nullptr;
int* tmp = new int[r - l + 1];
if (l == r) {
tmp[0] = a[l];
return tmp;
}
int mid = l + ((r - l) >> 1);
int llen = mid - l + 1, rlen = r - mid;
int* la = merge(l, mid);
int* ra = merge(mid + 1, r);
int i = 0, j = 0, cnt = 0;
for (; i < llen && j < rlen; ) {
if (la[i] > ra[j]) {
tmp[cnt ++] = ra[j ++];
} else {
tmp[cnt ++] = la[i ++];
}
}
// 上边的循环结束之后,可能存在一个数组还未完全遍历。
while(i < llen) tmp[cnt ++] = la[i ++];
while(j < rlen) tmp[cnt ++] = ra[j ++];
return tmp;
}
Acwing 787. 归并排序
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
const int N = 100000 + 100;
int a[N], tmp[N];
void merge(int q[], int l, int r) {
if (l >= r) return;
int mid = l + ((r - l) >> 1);
merge(q, l, mid);
merge(q, mid + 1, r);
int i = l, j = mid + 1, cnt = 0;
for (; i <= mid && j <= r; ) {
if (q[i] > q[j]) {
tmp[cnt ++] = q[j ++];
} else {
tmp[cnt ++] = q[i ++];
}
}
while(i <= mid) tmp[cnt ++] = q[i ++];
while(j <= r) tmp[cnt ++] = q[j ++];
for (int i = l, j = 0; i <= r; i ++, j ++)
q[i] = tmp[j];
}
int main()
{
int n;
cin >> n;
for (int i = 0; i < n; i ++) cin >> a[i];
merge(a, 0, n - 1);
for (int i = 0; i < n; i ++) {
printf("%d ", a[i]);
}
printf("\n");
}
Acwing 788. 逆序对的数量
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N = 100100;
int a[N];
int n;
LL ans;
void print_arr(int* arr, int size) {
for (int i = 0; i < size; i ++) {
printf("%d ", arr[i]);
}
printf("\n");
}
int* merge(int l, int r) {
if (l > r) return nullptr;
int* tmp = new int[r - l + 1];
if (r == l) {
tmp[0] = a[l];
return tmp;
}
int mid = l + r >> 1;
int* larr = merge(l, mid);
int* rarr = merge(mid + 1, r);
int llen = (mid - l) + 1, rlen = (r - mid - 1) + 1;
// printf("l:\n");
// print_arr(larr, llen);
// printf("r:\n");
// print_arr(rarr, rlen);
int i = 0, j = 0, cnt = 0;
for (; i < llen && j < rlen;)
{
if (larr[i] > rarr[j]) {
ans += (llen - 1 - i) + 1;
tmp[cnt ++] = rarr[j ++];
} else {
tmp[cnt ++] = larr[i ++];
}
}
while(i < llen) tmp[cnt ++] = larr[i ++];
while(j < rlen) tmp[cnt ++] = rarr[j ++];
// printf("merge\n");
// print_arr(tmp, (r - l) + 1);
// printf("ans : %d\n", ans);
return tmp;
}
int main()
{
cin >> n;
for (int i = 0; i < n; i ++) cin >> a[i];
int* h = merge(0, n - 1);
// for (int i = 0; i < n; i ++) {
// printf("%d\n", h[i]);
// }
printf("%lld\n", ans);
return 0;
}
Leetcode 493. 翻转对
class Solution {
long ans = 0;
public int reversePairs(int[] nums) {
int n = nums.length;
mergeSort(nums, 0, n - 1);
return (int)ans;
}
void mergeSort(int[] nums, int l, int r) {
if (l >= r) return;
int[] tmp = new int[r - l + 1];
int mid = l + ((r - l) >> 1);
mergeSort(nums, l, mid);
mergeSort(nums, mid + 1, r);
int i = l, j = mid + 1, cnt = 0;
int base = 0;
for (; i <= mid; i ++) {
while (j <= r && (long)nums[i] > 2L * nums[j]) {
j ++;
}
ans += (j - (mid + 1));
}
i = l;
j = mid + 1;
for (; i <= mid && j <= r; ) {
if (nums[i] > nums[j]) tmp[cnt ++] = nums[j ++];
else tmp[cnt ++] = nums[i ++];
}
while(i <= mid) tmp[cnt ++] = nums[i ++];
while(j <= r) tmp[cnt ++] = nums[j ++];
for (int k = 0; k < cnt; k ++)
nums[l + k] = tmp[k];
}
}
Leetcode 315. 计算右侧小于当前元素的个数
- 这个题比较恶心的就是要维护元素原来的位置
class Node {
int x;
int id;
Node(int x, int id) {
this.x = x;
this.id = id;
}
}
class Solution {
List<Integer> ans = null;
public List<Integer> countSmaller(int[] nums) {
int n = nums.length;
ans = new ArrayList<>(Collections.nCopies(n, 0));
Node[] nodes = new Node[n];
for (int i = 0; i < n; i ++) {
nodes[i] = new Node(nums[i], i);
}
merge(nodes, 0, n - 1);
return ans;
}
void merge(Node[] nodes, int l, int r) {
if (l >= r) return;
Node[] tmp = new Node[r - l + 1];
int mid = l + ((r - l) >> 1);
merge(nodes, l, mid);
merge(nodes, mid + 1, r);
int i = l, j = mid + 1, cnt = 0;
int base = 0;
for (; i <= mid;) {
if (j == r + 1 || nodes[i].x <= nodes[j].x) {
ans.set(nodes[i].id, ans.get(nodes[i].id) + base);
tmp[cnt ++] = nodes[i ++];
} else {
tmp[cnt ++] = nodes[j ++];
base ++;
}
}
while (j <= r) tmp[cnt ++] = nodes[j ++];
for (int k = 0; k < cnt; k ++)
nodes[l + k] = tmp[k];
}
}
Leetcode 327. 区间和的个数(前缀和)
- 这个题首先要想到利用前缀和将原来的数组进行转换。
- 要求的是区间和属于[lower, upper]区间的个数,转化为数学符号之后就是这样:
l
o
w
e
r
<
=
s
u
m
[
i
]
−
s
u
m
[
j
]
<
=
u
p
p
e
r
lower <= sum[i] - sum[j] <= upper
lower<=sum[i]−sum[j]<=upper
- 对于这样的不等式,可以分两步来考虑:
- 将连续不等式拆分成单个不等式, s u m [ i ] − s u m [ j ] < = u p p e r sum[i] - sum[j] <= upper sum[i]−sum[j]<=upper
- 将变量
i
固定,求另外一个变量的值
- 之后对于刚才的连续不等式就可以计算出符合条件的区间
[m, n]
- 对于这样的不等式,可以分两步来考虑:
- 计算符合条件的区间的时机:在合并阶段,
i
的范围是[mid + 1, r]
class Solution {
int lower = 0, upper = 0, ans = 0;
public int countRangeSum(int[] nums, int lower, int upper) {
this.upper = upper;
this.lower = lower;
int n = nums.length;
long[] pre = new long[n + 1];
for (int i = 1; i <= n; i ++)
pre[i] = pre[i - 1] + nums[i - 1];
merge(pre, 0, n);
return ans;
}
void merge(long[] nums, int l, int r) {
if (l >= r) return;
long[] tmp = new long[r - l + 1];
int mid = l + ((r - l) >> 1);
merge(nums, l, mid);
merge(nums, mid + 1, r);
// 核心代码
for (int i = mid + 1, j = l, k = l; i <= r; i ++) {
while (j <= mid && nums[i] - nums[j] > upper) j ++;
while (k <= mid && nums[i] - nums[k] >= lower) k ++;
ans += k - j;
}
int cnt = 0;
for (int i = l, j = mid + 1; i <= mid || j <= r; ) {
if (i == mid + 1) tmp[cnt ++] = nums[j ++];
else if (j == r + 1) tmp[cnt ++] = nums[i ++];
else {
if (nums[i] > nums[j])
tmp[cnt ++] = nums[j ++];
else
tmp[cnt ++] = nums[i ++];
}
}
for (int i = 0; i < cnt; i ++)
nums[i + l] = tmp[i];
}
}
Acwing 65. 数组中的逆序对
- 题意就在题面上,所以直接套模板。
class Solution {
int ans = 0;
public int inversePairs(int[] nums) {
int n = nums.length;
mergeSort(nums, 0, n - 1);
return ans;
}
void mergeSort(int[] nums, int l, int r) {
if (l >= r) return;
int[] tmp = new int[r - l + 1];
int mid = l + ((r - l) >> 1);
mergeSort(nums, l, mid);
mergeSort(nums, mid + 1, r);
int i = l, j = mid + 1, cnt = 0;
for (; i <= mid && j <= r; ) {
if (nums[i] > nums[j]) {
ans += (mid - i) + 1;
tmp[cnt ++] = nums[j ++];
} else {
tmp[cnt ++] = nums[i ++];
}
}
while (i <= mid) tmp[cnt ++] = nums[i ++];
while (j <= r) tmp[cnt ++] = nums[j ++];
for (int k = 0; k < cnt; k ++)
nums[l + k] = tmp[k];
}
}
Acwing 107. 超快速排序
- 根据题意可以分析出本题是要求逆序的数量, 那就直接套模板。
import java.util.Scanner;
class Main {
static long ans = 0;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = 0;
while((n = sc.nextInt()) != 0) {
ans = 0;
int[] nums = new int[n];
for (int i = 0; i < n; i ++) {
nums[i] = sc.nextInt();
}
mergeSort(nums, 0, n - 1);
System.out.println(ans);
}
}
static void mergeSort(int[] nums, int l, int r) {
if (l >= r) return;
int[] tmp = new int[r - l + 1];
int mid = l + ((r - l) >> 1);
mergeSort(nums, l, mid);
mergeSort(nums, mid + 1, r);
int i = l, j = mid + 1, cnt = 0;
for (; i <= mid && j <= r; ) {
if (nums[i] > nums[j]) {
ans += (mid - i) + 1;
tmp[cnt ++] = nums[j ++];
} else {
tmp[cnt ++] = nums[i ++];
}
}
while (i <= mid) tmp[cnt ++] = nums[i ++];
while (j <= r) tmp[cnt ++] = nums[j ++];
for (int k = 0; k < cnt; k ++)
nums[l + k] = tmp[k];
}
}