题目描述
给你一个整数数组 nums ，按要求返回一个新数组 counts 。数组 counts 有该性质： counts[i] 的值是 nums[i] 右侧小于 nums[i] 的元素的数量。
示例 1：
输入：nums = [5,2,6,1]
输出：[2,1,1,0]
解释：
5 的右侧有 2 个更小的元素 (2 和 1)
2 的右侧仅有 1 个更小的元素 (1)
6 的右侧有 1 个更小的元素 (1)
1 的右侧有 0 个更小的元素
示例 2：
输入：nums = [-1]
输出：[0]
示例 3：
输入：nums = [-1,-1]
输出：[0,0]
提示：
1 <= nums.length <= 105-104 <= nums[i] <= 104
解法
方法一：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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33 | 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 countSmaller(self, nums: List[int]) -> List[int]:
alls = sorted(set(nums))
m = {v: i for i, v in enumerate(alls, 1)}
tree = BinaryIndexedTree(len(m))
ans = []
for v in nums[::-1]:
x = m[v]
tree.update(x, 1)
ans.append(tree.query(x - 1))
return ans[::-1]
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53 | class Solution {
public List<Integer> countSmaller(int[] nums) {
Set<Integer> s = new HashSet<>();
for (int v : nums) {
s.add(v);
}
List<Integer> alls = new ArrayList<>(s);
alls.sort(Comparator.comparingInt(a -> a));
int n = alls.size();
Map<Integer, Integer> m = new HashMap<>(n);
for (int i = 0; i < n; ++i) {
m.put(alls.get(i), i + 1);
}
BinaryIndexedTree tree = new BinaryIndexedTree(n);
LinkedList<Integer> ans = new LinkedList<>();
for (int i = nums.length - 1; i >= 0; --i) {
int x = m.get(nums[i]);
tree.update(x, 1);
ans.addFirst(tree.query(x - 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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49 | 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:
vector<int> countSmaller(vector<int>& nums) {
unordered_set<int> s(nums.begin(), nums.end());
vector<int> alls(s.begin(), s.end());
sort(alls.begin(), alls.end());
unordered_map<int, int> m;
int n = alls.size();
for (int i = 0; i < n; ++i) m[alls[i]] = i + 1;
BinaryIndexedTree* tree = new BinaryIndexedTree(n);
vector<int> ans(nums.size());
for (int i = nums.size() - 1; i >= 0; --i) {
int x = m[nums[i]];
tree->update(x, 1);
ans[i] = tree->query(x - 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 countSmaller(nums []int) []int {
s := make(map[int]bool)
for _, v := range nums {
s[v] = true
}
var alls []int
for v := range s {
alls = append(alls, v)
}
sort.Ints(alls)
m := make(map[int]int)
for i, v := range alls {
m[v] = i + 1
}
ans := make([]int, len(nums))
tree := newBinaryIndexedTree(len(alls))
for i := len(nums) - 1; i >= 0; i-- {
x := m[nums[i]]
tree.update(x, 1)
ans[i] = tree.query(x - 1)
}
return ans
}
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方法二：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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58 | 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(n << 2)]
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 += v
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 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
def pushup(self, u):
self.tr[u].v = self.tr[u << 1].v + self.tr[u << 1 | 1].v
class Solution:
def countSmaller(self, nums: List[int]) -> List[int]:
s = sorted(set(nums))
m = {v: i for i, v in enumerate(s, 1)}
tree = SegmentTree(len(s))
ans = []
for v in nums[::-1]:
x = m[v]
ans.append(tree.query(1, 1, x - 1))
tree.modify(1, x, 1)
return ans[::-1]
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85 | class Solution {
public List<Integer> countSmaller(int[] nums) {
Set<Integer> s = new HashSet<>();
for (int v : nums) {
s.add(v);
}
List<Integer> alls = new ArrayList<>(s);
alls.sort(Comparator.comparingInt(a -> a));
int n = alls.size();
Map<Integer, Integer> m = new HashMap<>(n);
for (int i = 0; i < n; ++i) {
m.put(alls.get(i), i + 1);
}
SegmentTree tree = new SegmentTree(n);
LinkedList<Integer> ans = new LinkedList<>();
for (int i = nums.length - 1; i >= 0; --i) {
int x = m.get(nums[i]);
tree.modify(1, x, 1);
ans.addFirst(tree.query(1, 1, x - 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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72 | 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:
vector<int> countSmaller(vector<int>& nums) {
unordered_set<int> s(nums.begin(), nums.end());
vector<int> alls(s.begin(), s.end());
sort(alls.begin(), alls.end());
unordered_map<int, int> m;
int n = alls.size();
for (int i = 0; i < n; ++i) m[alls[i]] = i + 1;
SegmentTree* tree = new SegmentTree(n);
vector<int> ans(nums.size());
for (int i = nums.size() - 1; i >= 0; --i) {
int x = m[nums[i]];
tree->modify(1, x, 1);
ans[i] = tree->query(1, 1, x - 1);
}
return ans;
}
};
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56 | type Pair struct {
val int
index int
}
var (
tmp []Pair
count []int
)
func countSmaller(nums []int) []int {
tmp, count = make([]Pair, len(nums)), make([]int, len(nums))
array := make([]Pair, len(nums))
for i, v := range nums {
array[i] = Pair{val: v, index: i}
}
sorted(array, 0, len(array)-1)
return count
}
func sorted(arr []Pair, low, high int) {
if low >= high {
return
}
mid := low + (high-low)/2
sorted(arr, low, mid)
sorted(arr, mid+1, high)
merge(arr, low, mid, high)
}
func merge(arr []Pair, low, mid, high int) {
left, right := low, mid+1
idx := low
for left <= mid && right <= high {
if arr[left].val <= arr[right].val {
count[arr[left].index] += right - mid - 1
tmp[idx], left = arr[left], left+1
} else {
tmp[idx], right = arr[right], right+1
}
idx++
}
for left <= mid {
count[arr[left].index] += right - mid - 1
tmp[idx] = arr[left]
idx, left = idx+1, left+1
}
for right <= high {
tmp[idx] = arr[right]
idx, right = idx+1, right+1
}
// 排序
for i := low; i <= high; i++ {
arr[i] = tmp[i]
}
}
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方法三：归并排序