题目描述
给你一个整数数组 nums ，找到其中最长严格递增子序列的长度。
子序列 是由数组派生而来的序列，删除（或不删除）数组中的元素而不改变其余元素的顺序。例如，[3,6,2,7] 是数组 [0,3,1,6,2,2,7] 的子序列。
示例 1：
输入：nums = [10,9,2,5,3,7,101,18]
输出：4
解释：最长递增子序列是 [2,3,7,101]，因此长度为 4 。
示例 2：
输入：nums = [0,1,0,3,2,3]
输出：4
示例 3：
输入：nums = [7,7,7,7,7,7,7]
输出：1
提示：
1 <= nums.length <= 2500-104 <= nums[i] <= 104
进阶：
- 你能将算法的时间复杂度降低到
O(n log(n)) 吗?
解法
方法一：动态规划
我们定义 $f[i]$ 表示以 $nums[i]$ 结尾的最长递增子序列的长度，初始时 $f[i] = 1$，答案为 $f[i]$ 的最大值。
对于 $f[i]$，我们需要枚举 $0 \le j \lt i$，如果 $nums[j] \lt nums[i]$，则 $f[i] = \max(f[i], f[j] + 1)$。
最后的答案即为 $f[i]$ 的最大值。
时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为数组长度。
| class Solution:
def lengthOfLIS(self, nums: List[int]) -> int:
n = len(nums)
f = [1] * n
for i in range(1, n):
for j in range(i):
if nums[j] < nums[i]:
f[i] = max(f[i], f[j] + 1)
return max(f)
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17 | class Solution {
public int lengthOfLIS(int[] nums) {
int n = nums.length;
int[] f = new int[n];
Arrays.fill(f, 1);
int ans = 1;
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (nums[j] < nums[i]) {
f[i] = Math.max(f[i], f[j] + 1);
}
}
ans = Math.max(ans, f[i]);
}
return ans;
}
}
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15 | class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
int n = nums.size();
vector<int> f(n, 1);
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (nums[j] < nums[i]) {
f[i] = max(f[i], f[j] + 1);
}
}
}
return *max_element(f.begin(), f.end());
}
};
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17 | func lengthOfLIS(nums []int) int {
n := len(nums)
f := make([]int, n)
for i := range f {
f[i] = 1
}
ans := 1
for i := 1; i < n; i++ {
for j := 0; j < i; j++ {
if nums[j] < nums[i] {
f[i] = max(f[i], f[j]+1)
ans = max(ans, f[i])
}
}
}
return ans
}
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12 | function lengthOfLIS(nums: number[]): number {
const n = nums.length;
const f: number[] = new Array(n).fill(1);
for (let i = 1; i < n; ++i) {
for (let j = 0; j < i; ++j) {
if (nums[j] < nums[i]) {
f[i] = Math.max(f[i], f[j] + 1);
}
}
}
return Math.max(...f);
}
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14 | impl Solution {
pub fn length_of_lis(nums: Vec<i32>) -> i32 {
let n = nums.len();
let mut f = vec![1; n];
for i in 1..n {
for j in 0..i {
if nums[j] < nums[i] {
f[i] = f[i].max(f[j] + 1);
}
}
}
*f.iter().max().unwrap()
}
}
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方法二：离散化 + 树状数组
我们将数组中的元素离散化，然后使用树状数组维护不大于某个元素的最长递增子序列的长度。
遍历数组中的每个元素 $x$，将其离散化，然后在树状数组中查找不大于 $x-1$ 的最长递增子序列的长度 $t$，则 $x$ 的最长递增子序列的长度为 $t+1$，更新答案，并且更新树状数组中 $x$ 的最长递增子序列的长度。
遍历完数组中的所有元素，即可得到答案。
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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28 | class BinaryIndexedTree:
def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] = max(self.c[x], v)
x += x & -x
def query(self, x: int) -> int:
mx = 0
while x:
mx = max(mx, self.c[x])
x -= x & -x
return mx
class Solution:
def lengthOfLIS(self, nums: List[int]) -> int:
s = sorted(set(nums))
m = len(s)
tree = BinaryIndexedTree(m)
for x in nums:
x = bisect_left(s, x) + 1
t = tree.query(x - 1) + 1
tree.update(x, t)
return tree.query(m)
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59 | class Solution {
public int lengthOfLIS(int[] nums) {
int[] s = nums.clone();
Arrays.sort(s);
int m = 0;
int n = s.length;
for (int i = 0; i < n; ++i) {
if (i == 0 || s[i] != s[i - 1]) {
s[m++] = s[i];
}
}
BinaryIndexedTree tree = new BinaryIndexedTree(m);
for (int x : nums) {
x = search(s, x, m);
int t = tree.query(x - 1) + 1;
tree.update(x, t);
}
return tree.query(m);
}
private int search(int[] nums, int x, int r) {
int l = 0;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l + 1;
}
}
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 v) {
while (x <= n) {
c[x] = Math.max(c[x], v);
x += x & -x;
}
}
public int query(int x) {
int mx = 0;
while (x > 0) {
mx = Math.max(mx, c[x]);
x -= x & -x;
}
return mx;
}
}
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42 | class BinaryIndexedTree {
public:
BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) {}
void update(int x, int v) {
while (x <= n) {
c[x] = max(c[x], v);
x += x & -x;
}
}
int query(int x) {
int mx = 0;
while (x) {
mx = max(mx, c[x]);
x -= x & -x;
}
return mx;
}
private:
int n;
vector<int> c;
};
class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
vector<int> s = nums;
sort(s.begin(), s.end());
s.erase(unique(s.begin(), s.end()), s.end());
BinaryIndexedTree tree(s.size());
for (int x : nums) {
x = lower_bound(s.begin(), s.end(), x) - s.begin() + 1;
int t = tree.query(x - 1) + 1;
tree.update(x, t);
}
return tree.query(s.size());
}
};
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45 | type BinaryIndexedTree struct {
n int
c []int
}
func newBinaryIndexedTree(n int) *BinaryIndexedTree {
return &BinaryIndexedTree{n, make([]int, n+1)}
}
func (bit *BinaryIndexedTree) update(x, v int) {
for x <= bit.n {
bit.c[x] = max(bit.c[x], v)
x += x & -x
}
}
func (bit *BinaryIndexedTree) query(x int) int {
mx := 0
for x > 0 {
mx = max(mx, bit.c[x])
x -= x & -x
}
return mx
}
func lengthOfLIS(nums []int) int {
n := len(nums)
s := make([]int, n)
copy(s, nums)
sort.Ints(s)
m := 0
for i, x := range s {
if i == 0 || x != s[i-1] {
s[m] = x
m++
}
}
tree := newBinaryIndexedTree(m)
for _, x := range nums {
x = sort.SearchInts(s[:m], x) + 1
t := tree.query(x-1) + 1
tree.update(x, t)
}
return tree.query(m)
}
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51 | class BinaryIndexedTree {
private n: number;
private c: number[];
constructor(n: number) {
this.n = n;
this.c = new Array(n + 1).fill(0);
}
update(x: number, v: number) {
while (x <= this.n) {
this.c[x] = Math.max(this.c[x], v);
x += x & -x;
}
}
query(x: number): number {
let mx = 0;
while (x) {
mx = Math.max(mx, this.c[x]);
x -= x & -x;
}
return mx;
}
}
function lengthOfLIS(nums: number[]): number {
const s = [...new Set(nums)].sort((a, b) => a - b);
const m = s.length;
const tree = new BinaryIndexedTree(m);
for (let x of nums) {
x = search(s, x);
const t = tree.query(x - 1) + 1;
tree.update(x, t);
}
return tree.query(m);
}
function search(nums: number[], x: number): number {
let l = 0,
r = nums.length - 1;
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l + 1;
}
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