673. 最长递增子序列的个数

题目描述

给定一个未排序的整数数组 nums ， 返回最长递增子序列的个数 。

注意这个数列必须是严格递增的。

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。

提示:

1 <= nums.length <= 2000
-10⁶ <= nums[i] <= 10⁶

解法

这是最长递增子序列的变形题。

方法一：动态规划

除了原有的 dp 数组之外，另加了 cnt 数组记录以 nums[i] 结尾的最长子序列的个数。

时间复杂度 $O(n^2)$。

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 $O(\log n)$。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。

本题我们使用树状数组 tree[x] 来维护以 x 结尾的最长上升子序列的长度，以及该长度对应的子序列个数。

时间复杂度 $O(n\log n)$。

Python3

class Solution:
    def findNumberOfLIS(self, nums: List[int]) -> int:
        maxLen, ans, n = 0, 0, len(nums)
        dp, cnt = [1] * n, [1] * n
        for i in range(n):
            for j in range(i):
                if nums[i] > nums[j]:
                    if dp[j] + 1 > dp[i]:
                        dp[i] = dp[j] + 1
                        cnt[i] = cnt[j]
                    elif dp[j] + 1 == dp[i]:
                        cnt[i] += cnt[j]
            if dp[i] > maxLen:
                maxLen = dp[i]
                ans = cnt[i]
            elif dp[i] == maxLen:
                ans += cnt[i]
        return ans

class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)
        self.d = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, val, cnt):
        while x <= self.n:
            if self.c[x] < val:
                self.c[x] = val
                self.d[x] = cnt
            elif self.c[x] == val:
                self.d[x] += cnt
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        val = cnt = 0
        while x:
            if self.c[x] > val:
                val = self.c[x]
                cnt = self.d[x]
            elif self.c[x] == val:
                cnt += self.d[x]
            x -= BinaryIndexedTree.lowbit(x)
        return val, cnt


class Solution:
    def findNumberOfLIS(self, nums: List[int]) -> int:
        s = sorted(set(nums))
        m = {v: i for i, v in enumerate(s, 1)}
        n = len(m)
        tree = BinaryIndexedTree(n)
        ans = 0
        for v in nums:
            x = m[v]
            val, cnt = tree.query(x - 1)
            tree.update(x, val + 1, max(cnt, 1))
        return tree.query(n)[1]

Java

class Solution {
    public int findNumberOfLIS(int[] nums) {
        int maxLen = 0, ans = 0, n = nums.length;
        int[] dp = new int[n];
        int[] cnt = new int[n];
        for (int i = 0; i < n; i++) {
            dp[i] = 1;
            cnt[i] = 1;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    if (dp[j] + 1 > dp[i]) {
                        dp[i] = dp[j] + 1;
                        cnt[i] = cnt[j];
                    } else if (dp[j] + 1 == dp[i]) {
                        cnt[i] += cnt[j];
                    }
                }
            }
            if (dp[i] > maxLen) {
                maxLen = dp[i];
                ans = cnt[i];
            } else if (dp[i] == maxLen) {
                ans += cnt[i];
            }
        }
        return ans;
    }
}

class Solution {
    public int findNumberOfLIS(int[] nums) {
        TreeSet<Integer> ts = new TreeSet();
        for (int v : nums) {
            ts.add(v);
        }
        int idx = 1;
        Map<Integer, Integer> m = new HashMap<>();
        for (int v : ts) {
            m.put(v, idx++);
        }
        int n = m.size();
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        for (int v : nums) {
            int x = m.get(v);
            int[] t = tree.query(x - 1);
            tree.update(x, t[0] + 1, Math.max(t[1], 1));
        }
        return tree.query(n)[1];
    }
}

class BinaryIndexedTree {
    private int n;
    private int[] c;
    private int[] d;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
        d = new int[n + 1];
    }

    public void update(int x, int val, int cnt) {
        while (x <= n) {
            if (c[x] < val) {
                c[x] = val;
                d[x] = cnt;
            } else if (c[x] == val) {
                d[x] += cnt;
            }
            x += lowbit(x);
        }
    }

    public int[] query(int x) {
        int val = 0;
        int cnt = 0;
        while (x > 0) {
            if (val < c[x]) {
                val = c[x];
                cnt = d[x];
            } else if (val == c[x]) {
                cnt += d[x];
            }
            x -= lowbit(x);
        }
        return new int[] {val, cnt};
    }

    public static int lowbit(int x) {
        return x & -x;
    }
}

C++

class Solution {
public:
    int findNumberOfLIS(vector<int>& nums) {
        int maxLen = 0, ans = 0, n = nums.size();
        vector<int> dp(n, 1), cnt(n, 1);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] > nums[j]) {
                    if (dp[j] + 1 > dp[i]) {
                        dp[i] = dp[j] + 1;
                        cnt[i] = cnt[j];
                    } else if (dp[j] + 1 == dp[i]) {
                        cnt[i] += cnt[j];
                    }
                }
            }
            if (dp[i] > maxLen) {
                maxLen = dp[i];
                ans = cnt[i];
            } else if (dp[i] == maxLen) {
                ans += cnt[i];
            }
        }
        return ans;
    }
};

class BinaryIndexedTree {
public:
    int n;
    vector<int> c;
    vector<int> d;

    BinaryIndexedTree(int _n)
        : n(_n)
        , c(_n + 1)
        , d(n + 1) {}

    void update(int x, int val, int cnt) {
        while (x <= n) {
            if (c[x] < val) {
                c[x] = val;
                d[x] = cnt;
            } else if (c[x] == val)
                d[x] += cnt;
            x += lowbit(x);
        }
    }

    vector<int> query(int x) {
        int val = 0, cnt = 0;
        while (x > 0) {
            if (val < c[x]) {
                val = c[x];
                cnt = d[x];
            } else if (val == c[x])
                cnt += d[x];
            x -= lowbit(x);
        }
        return {val, cnt};
    }

    int lowbit(int x) {
        return x & -x;
    }
};

class Solution {
public:
    int findNumberOfLIS(vector<int>& nums) {
        set<int> s(nums.begin(), nums.end());
        int idx = 1;
        unordered_map<int, int> m;
        for (int v : s) m[v] = idx++;
        int n = m.size();
        BinaryIndexedTree* tree = new BinaryIndexedTree(n);
        for (int v : nums) {
            int x = m[v];
            auto t = tree->query(x - 1);
            tree->update(x, t[0] + 1, max(t[1], 1));
        }
        return tree->query(n)[1];
    }
};

Go

func findNumberOfLIS(nums []int) int {
	maxLen, ans, n := 0, 0, len(nums)
	dp, cnt := make([]int, n), make([]int, n)
	for i := 0; i < n; i++ {
		dp[i] = 1
		cnt[i] = 1
		for j := 0; j < i; j++ {
			if nums[i] > nums[j] {
				if dp[j]+1 > dp[i] {
					dp[i] = dp[j] + 1
					cnt[i] = cnt[j]
				} else if dp[j]+1 == dp[i] {
					cnt[i] += cnt[j]
				}
			}
		}
		if dp[i] > maxLen {
			maxLen = dp[i]
			ans = cnt[i]
		} else if dp[i] == maxLen {
			ans += cnt[i]
		}
	}
	return ans
}

type BinaryIndexedTree struct {
	n int
	c []int
	d []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	d := make([]int, n+1)
	return &BinaryIndexedTree{n, c, d}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
	return x & -x
}

func (this *BinaryIndexedTree) update(x, val, cnt int) {
	for x <= this.n {
		if this.c[x] < val {
			this.c[x] = val
			this.d[x] = cnt
		} else if this.c[x] == val {
			this.d[x] += cnt
		}
		x += this.lowbit(x)
	}
}

func (this *BinaryIndexedTree) query(x int) []int {
	var val, cnt int
	for x > 0 {
		if val < this.c[x] {
			val = this.c[x]
			cnt = this.d[x]
		} else if val == this.c[x] {
			cnt += this.d[x]
		}
		x -= this.lowbit(x)
	}
	return []int{val, cnt}
}

func findNumberOfLIS(nums []int) int {
	s := make(map[int]bool)
	for _, v := range nums {
		s[v] = true
	}
	var t []int
	for v, _ := range s {
		t = append(t, v)
	}
	sort.Ints(t)
	m := make(map[int]int)
	for i, v := range t {
		m[v] = i + 1
	}
	n := len(m)
	tree := newBinaryIndexedTree(n)
	for _, v := range nums {
		x := m[v]
		t := tree.query(x - 1)
		tree.update(x, t[0]+1, max(t[1], 1))
	}
	return tree.query(n)[1]
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Rust

impl Solution {
    pub fn find_number_of_lis(nums: Vec<i32>) -> i32 {
        let mut max_len = 0;
        let mut ans = 0;
        let n = nums.len();
        let mut dp = vec![1; n];
        let mut cnt = vec![1; n];
        for i in 0..n {
            for j in 0..i {
                if nums[i] > nums[j] {
                    if dp[j] + 1 > dp[i] {
                        dp[i] = dp[j] + 1;
                        cnt[i] = cnt[j];
                    } else if dp[j] + 1 == dp[i] {
                        cnt[i] += cnt[j];
                    }
                }
            }
            if dp[i] > max_len {
                max_len = dp[i];
                ans = cnt[i];
            } else if dp[i] == max_len {
                ans += cnt[i];
            }
        }
        ans
    }
}

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673. 最长递增子序列的个数

题目描述

解法

Python3

Java

C++

Go

Rust

...

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673. 最长递增子序列的个数

题目描述

解法

Python3

Java

C++

Go

Rust

...