Question
Given an unsorted array of integers, find the number of longest increasing subsequence.
Example 1:
Input: [1,3,5,4,7] Output: 2 Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].
Example 2:
Input: [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.
Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.
Difficulty:Medium
Category:Dynamic-Programming
Analyze
Solution
Solution 1: 递归
参考博客:花花酱 LeetCode 673. Number of Longest Increasing
// Runtime: 80ms
class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
if (nums.empty()) return 0;
const int n = nums.size();
f_ = vector<int>(n, 0);
c_ = vector<int>(n, 0);
// max_len must be difined in local, it will changed in the count function
int max_len = 0;
for (int i = 0; i < n; ++i) max_len = max(max_len, LIS(nums, i));
int ans = 0;
for (int i = 0; i < n; ++i) {
if (LIS(nums, i) == max_len) {
ans += count(nums, i);
}
}
return ans;
}
private:
vector<int> c_;
vector<int> f_;
int count(const vector<int>& nums, int n) {
if (n == 0) return 1;
if (c_[n] > 0) return c_[n];
int max_len = LIS(nums, n);
int total_count = 0;
for (int i = 0; i < n; ++i) {
// 比最后一个元素大,并且长度正好大于1
if (nums[n] > nums[i] && LIS(nums, i) == max_len - 1) {
total_count += count(nums, i);
}
}
//如果数量为0,那就是在之前找不到一个子集,所以就只能使用这一个数字作为1,就是单个元素的情况
if (total_count == 0) {
total_count = 1;
}
c_[n] = total_count;
return c_[n];
}
int LIS(const vector<int>& nums, int r) {
if (r == 0) return 1;
if (f_[r] > 0) return f_[r];
int ans = 1;
for (int i = 0; i < r; ++i) {
if (nums[r] > nums[i]) {
ans = max(ans, LIS(nums, i) + 1);
}
}
f_[r] = ans;
return f_[r];
}
};
Solution 2: 动态规划
// Solution2: Dynamic-Programming
// Runtime: 32ms
class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
if (nums.empty()) return 0;
int n = nums.size();
vector<int> f(n, 1);
vector<int> c(n, 1);
// f[i] 表示前面的元素可以组合得到的最长的元素之和
for (int i = 1; i < n; ++i)
for (int j = 0; j < i; ++j)
// 最后一个元素大于前面的元素的话,如果子问题的长度+1比现在的长度大的话,那就表示找到了一个新的解
if (nums[i] > nums[j]) {
if (f[j] + 1 > f[i]) {
f[i] = f[j] + 1;
//新的解,但是数量却没有变化的
c[i] = c[j];
} else if (f[j] + 1 == f[i]) {
//之前的最长子序列加一的长度和现在的长度相等,那么就表示可以把之前的子序列的长度归入到这里面进行统计
c[i] += c[j];
}
}
int max_len = *std::max_element(f.begin(), f.end());
int ans = 0;
for (int i = 0; i < n; ++i) {
if (f[i] == max_len) ans += c[i];
}
return ans;
}
};