题目描述
给定一个数组 nums ，如果 i < j 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个重要翻转对。
你需要返回给定数组中的重要翻转对的数量。
示例 1:
输入: [1,3,2,3,1]
输出: 2
示例 2:
输入: [2,4,3,5,1]
输出: 3
注意:
- 给定数组的长度不会超过
50000。
- 输入数组中的所有数字都在32位整数的表示范围内。
解法
方法一：归并排序
归并排序的过程中，如果左边的数大于右边的数，则右边的数与左边的数之后的数都构成逆序对。
时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组长度。
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29 | class Solution:
def reversePairs(self, nums: List[int]) -> int:
def merge_sort(l, r):
if l >= r:
return 0
mid = (l + r) >> 1
ans = merge_sort(l, mid) + merge_sort(mid + 1, r)
t = []
i, j = l, mid + 1
while i <= mid and j <= r:
if nums[i] <= 2 * nums[j]:
i += 1
else:
ans += mid - i + 1
j += 1
i, j = l, mid + 1
while i <= mid and j <= r:
if nums[i] <= nums[j]:
t.append(nums[i])
i += 1
else:
t.append(nums[j])
j += 1
t.extend(nums[i : mid + 1])
t.extend(nums[j : r + 1])
nums[l : r + 1] = t
return ans
return merge_sort(0, len(nums) - 1)
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47 | class Solution {
private int[] nums;
private int[] t;
public int reversePairs(int[] nums) {
this.nums = nums;
int n = nums.length;
this.t = new int[n];
return mergeSort(0, n - 1);
}
private int mergeSort(int l, int r) {
if (l >= r) {
return 0;
}
int mid = (l + r) >> 1;
int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
int i = l, j = mid + 1, k = 0;
while (i <= mid && j <= r) {
if (nums[i] <= nums[j] * 2L) {
++i;
} else {
ans += mid - i + 1;
++j;
}
}
i = l;
j = mid + 1;
while (i <= mid && j <= r) {
if (nums[i] <= nums[j]) {
t[k++] = nums[i++];
} else {
t[k++] = nums[j++];
}
}
while (i <= mid) {
t[k++] = nums[i++];
}
while (j <= r) {
t[k++] = nums[j++];
}
for (i = l; i <= r; ++i) {
nums[i] = t[i - l];
}
return ans;
}
}
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43 | class Solution {
public:
int reversePairs(vector<int>& nums) {
int n = nums.size();
int t[n];
function<int(int, int)> mergeSort = [&](int l, int r) -> int {
if (l >= r) {
return 0;
}
int mid = (l + r) >> 1;
int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
int i = l, j = mid + 1, k = 0;
while (i <= mid && j <= r) {
if (nums[i] <= nums[j] * 2LL) {
++i;
} else {
ans += mid - i + 1;
++j;
}
}
i = l;
j = mid + 1;
while (i <= mid && j <= r) {
if (nums[i] <= nums[j]) {
t[k++] = nums[i++];
} else {
t[k++] = nums[j++];
}
}
while (i <= mid) {
t[k++] = nums[i++];
}
while (j <= r) {
t[k++] = nums[j++];
}
for (i = l; i <= r; ++i) {
nums[i] = t[i - l];
}
return ans;
};
return mergeSort(0, n - 1);
}
};
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42 | func reversePairs(nums []int) int {
n := len(nums)
t := make([]int, n)
var mergeSort func(l, r int) int
mergeSort = func(l, r int) int {
if l >= r {
return 0
}
mid := (l + r) >> 1
ans := mergeSort(l, mid) + mergeSort(mid+1, r)
i, j, k := l, mid+1, 0
for i <= mid && j <= r {
if nums[i] <= nums[j]*2 {
i++
} else {
ans += mid - i + 1
j++
}
}
i, j = l, mid+1
for i <= mid && j <= r {
if nums[i] <= nums[j] {
t[k] = nums[i]
k, i = k+1, i+1
} else {
t[k] = nums[j]
k, j = k+1, j+1
}
}
for ; i <= mid; i, k = i+1, k+1 {
t[k] = nums[i]
}
for ; j <= r; j, k = j+1, k+1 {
t[k] = nums[j]
}
for i = l; i <= r; i++ {
nums[i] = t[i-l]
}
return ans
}
return mergeSort(0, n-1)
}
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方法二：树状数组
树状数组，也称作“二叉索引树”（Binary Indexed Tree）或 Fenwick 树。 它可以高效地实现如下两个操作：
- 单点更新
update(x, delta)： 把序列 x 位置的数加上一个值 delta；
- 前缀和查询
query(x)：查询序列 [1,...x] 区间的区间和，即位置 x 的前缀和。
这两个操作的时间复杂度均为 $O(\log n)$。
树状数组最基本的功能就是求比某点 x 小的点的个数（这里的比较是抽象的概念，可以是数的大小、坐标的大小、质量的大小等等）。
比如给定数组 a[5] = {2, 5, 3, 4, 1}，求 b[i] = 位置 i 左边小于等于 a[i] 的数的个数。对于此例，b[5] = {0, 1, 1, 2, 0}。
解决方案是直接遍历数组，每个位置先求出 query(a[i])，然后再修改树状数组 update(a[i], 1) 即可。当数的范围比较大时，需要进行离散化，即先进行去重并排序，然后对每个数字进行编号。
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36 | class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
@staticmethod
def lowbit(x):
return x & -x
def update(self, x, delta):
while x <= self.n:
self.c[x] += delta
x += BinaryIndexedTree.lowbit(x)
def query(self, x):
s = 0
while x > 0:
s += self.c[x]
x -= BinaryIndexedTree.lowbit(x)
return s
class Solution:
def reversePairs(self, nums: List[int]) -> int:
s = set()
for num in nums:
s.add(num)
s.add(num * 2)
alls = sorted(s)
m = {v: i for i, v in enumerate(alls, 1)}
ans = 0
tree = BinaryIndexedTree(len(m))
for num in nums[::-1]:
ans += tree.query(m[num] - 1)
tree.update(m[num * 2], 1)
return ans
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52 | class Solution {
public int reversePairs(int[] nums) {
TreeSet<Long> ts = new TreeSet<>();
for (int num : nums) {
ts.add((long) num);
ts.add((long) num * 2);
}
Map<Long, Integer> m = new HashMap<>();
int idx = 0;
for (long num : ts) {
m.put(num, ++idx);
}
BinaryIndexedTree tree = new BinaryIndexedTree(m.size());
int ans = 0;
for (int i = nums.length - 1; i >= 0; --i) {
int x = m.get((long) nums[i]);
ans += tree.query(x - 1);
tree.update(m.get((long) nums[i] * 2), 1);
}
return ans;
}
}
class BinaryIndexedTree {
private int n;
private int[] c;
public BinaryIndexedTree(int n) {
this.n = n;
c = new int[n + 1];
}
public void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += lowbit(x);
}
}
public int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= lowbit(x);
}
return s;
}
public static int lowbit(int x) {
return x & -x;
}
}
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50 | class BinaryIndexedTree {
public:
int n;
vector<int> c;
BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) {}
void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += lowbit(x);
}
}
int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= lowbit(x);
}
return s;
}
int lowbit(int x) {
return x & -x;
}
};
class Solution {
public:
int reversePairs(vector<int>& nums) {
set<long long> s;
for (int num : nums) {
s.insert(num);
s.insert(num * 2ll);
}
unordered_map<long long, int> m;
int idx = 0;
for (long long num : s) m[num] = ++idx;
BinaryIndexedTree* tree = new BinaryIndexedTree(m.size());
int ans = 0;
for (int i = nums.size() - 1; i >= 0; --i) {
ans += tree->query(m[nums[i]] - 1);
tree->update(m[nums[i] * 2ll], 1);
}
return ans;
}
};
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53 | type BinaryIndexedTree struct {
n int
c []int
}
func newBinaryIndexedTree(n int) *BinaryIndexedTree {
c := make([]int, n+1)
return &BinaryIndexedTree{n, c}
}
func (this *BinaryIndexedTree) lowbit(x int) int {
return x & -x
}
func (this *BinaryIndexedTree) update(x, delta int) {
for x <= this.n {
this.c[x] += delta
x += this.lowbit(x)
}
}
func (this *BinaryIndexedTree) query(x int) int {
s := 0
for x > 0 {
s += this.c[x]
x -= this.lowbit(x)
}
return s
}
func reversePairs(nums []int) int {
s := make(map[int]bool)
for _, num := range nums {
s[num] = true
s[num*2] = true
}
var alls []int
for num := range s {
alls = append(alls, num)
}
sort.Ints(alls)
m := make(map[int]int)
for i, num := range alls {
m[num] = i + 1
}
tree := newBinaryIndexedTree(len(m))
ans := 0
for i := len(nums) - 1; i >= 0; i-- {
ans += tree.query(m[nums[i]] - 1)
tree.update(m[nums[i]*2], 1)
}
return ans
}
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方法三：线段树
线段树将整个区间分割为多个不连续的子区间，子区间的数量不超过 log(width)。更新某个元素的值，只需要更新 log(width) 个区间，并且这些区间都包含在一个包含该元素的大区间内。
- 线段树的每个节点代表一个区间；
- 线段树具有唯一的根节点，代表的区间是整个统计范围，如
[1, N]；
- 线段树的每个叶子节点代表一个长度为 1 的元区间
[x, x]；
- 对于每个内部节点
[l, r]，它的左儿子是 [l, mid]，右儿子是 [mid + 1, r], 其中 mid = ⌊(l + r) / 2⌋ (即向下取整)。
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62 | class Node:
def __init__(self):
self.l = 0
self.r = 0
self.v = 0
class SegmentTree:
def __init__(self, n):
self.tr = [Node() for _ in range(4 * n)]
self.build(1, 1, n)
def build(self, u, l, r):
self.tr[u].l = l
self.tr[u].r = r
if l == r:
return
mid = (l + r) >> 1
self.build(u << 1, l, mid)
self.build(u << 1 | 1, mid + 1, r)
def modify(self, u, x, v):
if self.tr[u].l == x and self.tr[u].r == x:
self.tr[u].v += 1
return
mid = (self.tr[u].l + self.tr[u].r) >> 1
if x <= mid:
self.modify(u << 1, x, v)
else:
self.modify(u << 1 | 1, x, v)
self.pushup(u)
def pushup(self, u):
self.tr[u].v = self.tr[u << 1].v + self.tr[u << 1 | 1].v
def query(self, u, l, r):
if self.tr[u].l >= l and self.tr[u].r <= r:
return self.tr[u].v
mid = (self.tr[u].l + self.tr[u].r) >> 1
v = 0
if l <= mid:
v += self.query(u << 1, l, r)
if r > mid:
v += self.query(u << 1 | 1, l, r)
return v
class Solution:
def reversePairs(self, nums: List[int]) -> int:
s = set()
for num in nums:
s.add(num)
s.add(num * 2)
alls = sorted(s)
m = {v: i for i, v in enumerate(alls, 1)}
tree = SegmentTree(len(m))
ans = 0
for v in nums[::-1]:
x = m[v]
ans += tree.query(1, 1, x - 1)
tree.modify(1, m[v * 2], 1)
return ans
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84 | class Solution {
public int reversePairs(int[] nums) {
TreeSet<Long> ts = new TreeSet<>();
for (int num : nums) {
ts.add((long) num);
ts.add((long) num * 2);
}
Map<Long, Integer> m = new HashMap<>();
int idx = 0;
for (long num : ts) {
m.put(num, ++idx);
}
SegmentTree tree = new SegmentTree(m.size());
int ans = 0;
for (int i = nums.length - 1; i >= 0; --i) {
int x = m.get((long) nums[i]);
ans += tree.query(1, 1, x - 1);
tree.modify(1, m.get((long) nums[i] * 2), 1);
}
return ans;
}
}
class Node {
int l;
int r;
int v;
}
class SegmentTree {
private Node[] tr;
public SegmentTree(int n) {
tr = new Node[4 * n];
for (int i = 0; i < tr.length; ++i) {
tr[i] = new Node();
}
build(1, 1, n);
}
public void build(int u, int l, int r) {
tr[u].l = l;
tr[u].r = r;
if (l == r) {
return;
}
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
}
public void modify(int u, int x, int v) {
if (tr[u].l == x && tr[u].r == x) {
tr[u].v += v;
return;
}
int mid = (tr[u].l + tr[u].r) >> 1;
if (x <= mid) {
modify(u << 1, x, v);
} else {
modify(u << 1 | 1, x, v);
}
pushup(u);
}
public void pushup(int u) {
tr[u].v = tr[u << 1].v + tr[u << 1 | 1].v;
}
public int query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) {
return tr[u].v;
}
int mid = (tr[u].l + tr[u].r) >> 1;
int v = 0;
if (l <= mid) {
v += query(u << 1, l, r);
}
if (r > mid) {
v += query(u << 1 | 1, l, r);
}
return v;
}
}
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73 | class Node {
public:
int l;
int r;
int v;
};
class SegmentTree {
public:
vector<Node*> tr;
SegmentTree(int n) {
tr.resize(4 * n);
for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
build(1, 1, n);
}
void build(int u, int l, int r) {
tr[u]->l = l;
tr[u]->r = r;
if (l == r) return;
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
}
void modify(int u, int x, int v) {
if (tr[u]->l == x && tr[u]->r == x) {
tr[u]->v += v;
return;
}
int mid = (tr[u]->l + tr[u]->r) >> 1;
if (x <= mid)
modify(u << 1, x, v);
else
modify(u << 1 | 1, x, v);
pushup(u);
}
void pushup(int u) {
tr[u]->v = tr[u << 1]->v + tr[u << 1 | 1]->v;
}
int query(int u, int l, int r) {
if (tr[u]->l >= l && tr[u]->r <= r) return tr[u]->v;
int mid = (tr[u]->l + tr[u]->r) >> 1;
int v = 0;
if (l <= mid) v = query(u << 1, l, r);
if (r > mid) v += query(u << 1 | 1, l, r);
return v;
}
};
class Solution {
public:
int reversePairs(vector<int>& nums) {
set<long long> s;
for (int num : nums) {
s.insert(num);
s.insert(num * 2ll);
}
unordered_map<long long, int> m;
int idx = 0;
for (long long num : s) m[num] = ++idx;
SegmentTree* tree = new SegmentTree(m.size());
int ans = 0;
for (int i = nums.size() - 1; i >= 0; --i) {
ans += tree->query(1, 1, m[nums[i]] - 1);
tree->modify(1, m[nums[i] * 2ll], 1);
}
return ans;
}
};
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