feat: add solutions to lc problem: No.2915 #1905

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Expand Up		@@ -52,34 +52,231 @@

		<!-- 这里可写通用的实现逻辑 -->

	方法一:动态规划

	我们定义 $f[i][j]$ 表示前 $i$ 个数中选取若干个数,使得这若干个数的和恰好为 $j$ 的最长子序列的长度。初始时 $f[0][0]=0,ドル其余位置均为 $-\infty$。

	对于 $f[i][j],ドル我们考虑第 $i$ 个数 $x,ドル如果不选取 $x,ドル那么 $f[i][j]=f[i-1][j]$;如果选取 $x,ドル那么 $f[i][j]=f[i-1][j-x]+1,ドル其中 $j\ge x$。因此我们有状态转移方程:

	$$
	f[i][j]=\max\{f[i-1][j],f[i-1][j-x]+1\}
	$$

	最终答案为 $f[n][target],ドル如果 $f[n][target]\le0,ドル则不存在和为 $target$ 的子序列,返回 $-1$。

	时间复杂度 $O(n\times target),ドル空间复杂度 $O(n\times target)$。其中 $n$ 为数组长度,而 $target$ 为目标值。

	我们注意到 $f[i][j]$ 的状态只与 $f[i-1][\cdot]$ 有关,因此我们可以优化掉第一维,将空间复杂度优化到 $O(target)$。

		<!-- tabs:start -->

		### Python3

		<!-- 这里可写当前语言的特殊实现逻辑 -->

		```python
	class Solution:
	def lengthOfLongestSubsequence(self, nums: List[int], target: int) -> int:
	n = len(nums)
	f = [[-inf] * (target + 1) for _ in range(n + 1)]
	f[0][0] = 0
	for i, x in enumerate(nums, 1):
	for j in range(target + 1):
	f[i][j] = f[i - 1][j]
	if j >= x:
	f[i][j] = max(f[i][j], f[i - 1][j - x] + 1)
	return -1 if f[n][target] <= 0 else f[n][target]
	```

	```python
	class Solution:
	def lengthOfLongestSubsequence(self, nums: List[int], target: int) -> int:
	f = [0] + [-inf] * target
	for x in nums:
	for j in range(target, x - 1, -1):
	f[j] = max(f[j], f[j - x] + 1)
	return -1 if f[-1] <= 0 else f[-1]
		```

		### Java

		<!-- 这里可写当前语言的特殊实现逻辑 -->

		```java
	class Solution {
	public int lengthOfLongestSubsequence(List<Integer> nums, int target) {
	int n = nums.size();
	int[][] f = new int[n + 1][target + 1];
	final int inf = 1 << 30;
	for (int[] g : f) {
	Arrays.fill(g, -inf);
	}
	f[0][0] = 0;
	for (int i = 1; i <= n; ++i) {
	int x = nums.get(i - 1);
	for (int j = 0; j <= target; ++j) {
	f[i][j] = f[i - 1][j];
	if (j >= x) {
	f[i][j] = Math.max(f[i][j], f[i - 1][j - x] + 1);
	}
	}
	}
	return f[n][target] <= 0 ? -1 : f[n][target];
	}
	}
	```

	```java
	class Solution {
	public int lengthOfLongestSubsequence(List<Integer> nums, int target) {
	int[] f = new int[target + 1];
	final int inf = 1 << 30;
	Arrays.fill(f, -inf);
	f[0] = 0;
	for (int x : nums) {
	for (int j = target; j >= x; --j) {
	f[j] = Math.max(f[j], f[j - x] + 1);
	}
	}
	return f[target] <= 0 ? -1 : f[target];
	}
	}
		```

		### C++

		```cpp
	class Solution {
	public:
	int lengthOfLongestSubsequence(vector<int>& nums, int target) {
	int n = nums.size();
	int f[n + 1][target + 1];
	memset(f, -0x3f, sizeof(f));
	f[0][0] = 0;
	for (int i = 1; i <= n; ++i) {
	int x = nums[i - 1];
	for (int j = 0; j <= target; ++j) {
	f[i][j] = f[i - 1][j];
	if (j >= x) {
	f[i][j] = max(f[i][j], f[i - 1][j - x] + 1);
	}
	}
	}
	return f[n][target] <= 0 ? -1 : f[n][target];
	}
	};
	```

	```cpp
	class Solution {
	public:
	int lengthOfLongestSubsequence(vector<int>& nums, int target) {
	int f[target + 1];
	memset(f, -0x3f, sizeof(f));
	f[0] = 0;
	for (int x : nums) {
	for (int j = target; j >= x; --j) {
	f[j] = max(f[j], f[j - x] + 1);
	}
	}
	return f[target] <= 0 ? -1 : f[target];
	}
	};
		```

		### Go

		```go
	func lengthOfLongestSubsequence(nums []int, target int) int {
	n := len(nums)
	f := make([][]int, n+1)
	for i := range f {
	f[i] = make([]int, target+1)
	for j := range f[i] {
	f[i][j] = -(1 << 30)
	}
	}
	f[0][0] = 0
	for i := 1; i <= n; i++ {
	x := nums[i-1]
	for j := 0; j <= target; j++ {
	f[i][j] = f[i-1][j]
	if j >= x {
	f[i][j] = max(f[i][j], f[i-1][j-x]+1)
	}
	}
	}
	if f[n][target] <= 0 {
	return -1
	}
	return f[n][target]
	}

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

	```go
	func lengthOfLongestSubsequence(nums []int, target int) int {
	f := make([]int, target+1)
	for i := range f {
	f[i] = -(1 << 30)
	}
	f[0] = 0
	for _, x := range nums {
	for j := target; j >= x; j-- {
	f[j] = max(f[j], f[j-x]+1)
	}
	}
	if f[target] <= 0 {
	return -1
	}
	return f[target]
	}

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

	### TypeScript

	```ts
	function lengthOfLongestSubsequence(nums: number[], target: number): number {
	const n = nums.length;
	const f: number[][] = Array.from({ length: n + 1 }, () => Array(target + 1).fill(-Infinity));
	f[0][0] = 0;
	for (let i = 1; i <= n; ++i) {
	const x = nums[i - 1];
	for (let j = 0; j <= target; ++j) {
	f[i][j] = f[i - 1][j];
	if (j >= x) {
	f[i][j] = Math.max(f[i][j], f[i - 1][j - x] + 1);
	}
	}
	}
	return f[n][target] <= 0 ? -1 : f[n][target];
	}
	```

	```ts
	function lengthOfLongestSubsequence(nums: number[], target: number): number {
	const f: number[] = Array(target + 1).fill(-Infinity);
	f[0] = 0;
	for (const x of nums) {
	for (let j = target; j >= x; --j) {
	f[j] = Math.max(f[j], f[j - x] + 1);
	}
	}
	return f[target] <= 0 ? -1 : f[target];
	}
		```

		### ...
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