673 Number of Longest Increasing Subsequence 最长递增子序列的个数
Description:
Given an integer array nums, return the number of longest increasing subsequences.
Notice that the sequence has to be strictly increasing.
Example:
Example 1:
Input: nums = [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
Example 2:
Input: nums = [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.
Constraints:
1 <= nums.length <= 2000
-10^6 <= nums[i] <= 10^6
题目描述:
给定一个未排序的整数数组，找到最长递增子序列的个数。
示例 :
示例 1:
输入: [1,3,5,4,7]
输出: 2
解释: 有两个最长递增子序列，分别是 [1, 3, 4, 7] 和[1, 3, 5, 7]。
示例 2:
输入: [2,2,2,2,2]
输出: 5
解释: 最长递增子序列的长度是1，并且存在5个子序列的长度为1，因此输出5。
注意:
给定的数组长度不超过 2000 并且结果一定是32位有符号整数。
思路:
- 动态规划
dp[i] 表示以 nums[i] 结尾的最长递增子序列的长度
count[i] 表示以 nums[i] 结尾的最长递增子序列的个数
dp[i] = dp[j] + 1, if nums[j] < nums[i] and dp[j] >= dp[i]
count[i] = count[j], if nums[j] < nums[i] and dp[j] >= dp[i]
count[i] += count[j], if nums[j] < nums[i] and dp[j] + 1 == dp[i]
其中 -1 < j < i
时间复杂度 O(n ^ 2), 空间复杂度 O(n) - 线段树
线段树额外记录当前结点的最长递增子序列的长度和数量
时间复杂度 O(nlgn), 空间复杂度 O(n)
代码:
C++:
class Solution
{
public:
int findNumberOfLIS(vector& nums)
{
int n = nums.size(), result = 0, max_length = 0;
vector dp(n, 0), count(n, 1);
for (int i = 0; i < n; i++)
{
for (int j = 0; j < i; j++)
{
if (nums[j] < nums[i])
{
if (dp[i] == dp[j] + 1) count[i] += count[j];
else if (dp[j] >= dp[i])
{
dp[i] = dp[j] + 1;
count[i] = count[j];
}
}
}
}
max_length = *max_element(dp.begin(), dp.end());
for (int i = 0; i < n; i++) if (dp[i] == max_length) result += count[i];
return result;
}
};
Java:
class Node {
int rangeLeft, rangeRight;
Node left, right;
Value val;
public Node(int start, int end) {
rangeLeft = start;
rangeRight = end;
left = null;
right = null;
val = new Value(0, 1);
}
public int getMid() {
return rangeLeft + ((rangeRight - rangeLeft) >>> 1);
}
public Node getLeft() {
if (left == null) left = new Node(rangeLeft, getMid());
return left;
}
public Node getRight() {
if (right == null) right = new Node(getMid() + 1, rangeRight);
return right;
}
}
class Value {
int length;
int count;
public Value(int l, int c) {
length = l;
count = c;
}
}
class Solution {
public Value merge(Value v1, Value v2) {
if (v1.length == v2.length) {
if (v1.length == 0) return new Value(0, 1);
return new Value(v1.length, v1.count + v2.count);
}
return v1.length > v2.length ? v1 : v2;
}
public void insert(Node node, int key, Value val) {
if (node.rangeLeft == node.rangeRight) {
node.val = merge(val, node.val);
return;
} else if (key <= node.getMid()) insert(node.getLeft(), key, val);
else insert(node.getRight(), key, val);
node.val = merge(node.getLeft().val, node.getRight().val);
}
public Value query(Node node, int key) {
if (node.rangeRight <= key) return node.val;
else if (node.rangeLeft > key) return new Value(0, 1);
else return merge(query(node.getLeft(), key), query(node.getRight(), key));
}
public int findNumberOfLIS(int[] nums) {
if (nums.length == 0) return 0;
int min = nums[0], max = nums[0];
for (int num: nums) {
min = Math.min(min, num);
max = Math.max(max, num);
}
Node root = new Node(min, max);
for (int num: nums) {
Value v = query(root, num - 1);
insert(root, num, new Value(v.length + 1, v.count));
}
return root.val.count;
}
}
Python:
class Solution:
def findNumberOfLIS(self, nums):
n = len(nums)
if n <= 1:
return n
dp, count = [0] * n, [1] * n
for i, num in enumerate(nums):
for j in range(i):
if nums[j] < nums[i]:
if dp[j] >= dp[i]:
dp[i] = 1 + dp[j]
count[i] = count[j]
elif dp[j] + 1 == dp[i]:
count[i] += count[j]
return sum(c for i, c in enumerate(count) if dp[i] == max(dp))