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Commit f3c9f14

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feat: add solutions to lc/lcof2 problems
* Update solutions to lc No.0188.Best Time to Buy and Sell Stock IV, fix doocs#623 * Add solutions to lcof2 No.102.Target Sum
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‎lcof2/剑指 Offer II 102. 加减的目标值/README.md‎

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<p><meta charset="UTF-8" />注意:本题与主站 494&nbsp;题相同:&nbsp;<a href="https://leetcode-cn.com/problems/target-sum/">https://leetcode-cn.com/problems/target-sum/</a></p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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题目可以转换为 `0-1` 背包问题,只不过下标可能会出现负数,需要特殊处理。
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也可以用 DFS 记忆化搜索。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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**0-1 背包**
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```python
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class Solution:
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def findTargetSumWays(self, nums: List[int], target: int) -> int:
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if target < -1000 or target > 1000:
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return 0
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n = len(nums)
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dp = [[0] * 2001 for i in range(n)]
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dp[0][nums[0] + 1000] += 1
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dp[0][-nums[0] + 1000] += 1
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for i in range(1, n):
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for j in range(-1000, 1001):
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if dp[i - 1][j + 1000] > 0:
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dp[i][j + nums[i] + 1000] += dp[i - 1][j + 1000]
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dp[i][j - nums[i] + 1000] += dp[i - 1][j + 1000]
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return dp[n - 1][target + 1000]
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```
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设:添加 `-` 号的元素之和为 `x`,则添加 `+` 号的元素之和为 `s - x`,`s - x - x = target`,`2x = s - target`。需要满足 `s - target` 一定大于等于 0,并且能够被 2 整除。
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```python
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class Solution:
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def findTargetSumWays(self, nums: List[int], target: int) -> int:
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s = sum(nums)
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if s - target < 0 or (s - target) % 2 != 0:
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return 0
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target = (s - target) // 2 + 1
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n = len(nums) + 1
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dp = [[0] * target for _ in range(n)]
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dp[0][0] = 1
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for i in range(1, n):
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for j in range(target):
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dp[i][j] = dp[i - 1][j]
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if nums[i - 1] <= j:
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dp[i][j] += dp[i - 1][j - nums[i - 1]]
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return dp[-1][-1]
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```
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空间优化:
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```python
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class Solution:
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def findTargetSumWays(self, nums: List[int], target: int) -> int:
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s = sum(nums)
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if s - target < 0 or (s - target) % 2 != 0:
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return 0
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target = (s - target) // 2 + 1
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n = len(nums) + 1
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dp = [0] * target
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dp[0] = 1
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for i in range(1, n):
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for j in range(target - 1, nums[i - 1] - 1, -1):
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dp[j] += dp[j - nums[i - 1]]
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return dp[-1]
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```
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**DFS**
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```python
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class Solution:
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def findTargetSumWays(self, nums: List[int], target: int) -> int:
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@lru_cache(None)
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def dfs(i, t):
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if i == n:
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if t == target:
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return 1
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return 0
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return dfs(i + 1, t + nums[i]) + dfs(i + 1, t - nums[i])
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ans, n = 0, len(nums)
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return dfs(0, 0)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int findTargetSumWays(int[] nums, int target) {
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if (target < -1000 || target > 1000) {
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return 0;
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}
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int n = nums.length;
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int[][] dp = new int[n][2001];
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dp[0][nums[0] + 1000] += 1;
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dp[0][-nums[0] + 1000] += 1;
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for (int i = 1; i < n; i++) {
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for (int j = -1000; j <= 1000; j++) {
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if (dp[i - 1][j + 1000] > 0) {
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dp[i][j + nums[i] + 1000] += dp[i - 1][j + 1000];
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dp[i][j - nums[i] + 1000] += dp[i - 1][j + 1000];
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}
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}
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}
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return dp[n - 1][target + 1000];
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}
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}
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```
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空间优化:
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```java
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class Solution {
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public int findTargetSumWays(int[] nums, int target) {
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int s = 0;
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for (int x : nums) {
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s += x;
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}
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if (s - target < 0 || (s - target) % 2 != 0) {
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return 0;
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}
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target = (s - target) / 2 + 1;
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int[] dp = new int[target];
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dp[0] = 1;
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for (int i = 1; i < nums.length + 1; ++i) {
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for (int j = target - 1; j >= nums[i - 1]; --j) {
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dp[j] += dp[j - nums[i - 1]];
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}
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}
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return dp[target - 1];
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}
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}
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```
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### **C++**
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空间优化:
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```cpp
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class Solution {
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public:
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int findTargetSumWays(vector<int>& nums, int target) {
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int s = 0;
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for (int x : nums) s += x;
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if (s - target < 0 || (s - target) % 2 != 0) return 0;
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target = (s - target) / 2 + 1;
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vector<int> dp(target);
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dp[0] = 1;
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for (int i = 1; i < nums.size() + 1; ++i)
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{
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for (int j = target - 1; j >= nums[i - 1]; --j)
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{
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dp[j] += dp[j - nums[i - 1]];
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}
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}
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return dp[target - 1];
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}
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};
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```
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### **Go**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```go
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func findTargetSumWays(nums []int, target int) int {
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if target < -1000 || target > 1000 {
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return 0
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}
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n := len(nums)
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dp := make([][]int, n)
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for i := 0; i < n; i++ {
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dp[i] = make([]int, 2001)
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}
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dp[0][nums[0]+1000] += 1
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dp[0][-nums[0]+1000] += 1
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for i := 1; i < n; i++ {
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for j := -1000; j <= 1000; j++ {
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if dp[i-1][j+1000] > 0 {
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dp[i][j+nums[i]+1000] += dp[i-1][j+1000]
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dp[i][j-nums[i]+1000] += dp[i-1][j+1000]
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}
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}
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}
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return dp[n-1][target+1000]
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}
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```
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空间优化:
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```go
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func findTargetSumWays(nums []int, target int) int {
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s := 0
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for _, x := range nums {
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s += x
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}
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if s-target < 0 || (s-target)%2 != 0 {
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return 0
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}
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target = (s-target)/2 + 1
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dp := make([]int, target)
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dp[0] = 1
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for i := 1; i < len(nums)+1; i++ {
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for j := target - 1; j >= nums[i-1]; j-- {
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dp[j] += dp[j-nums[i-1]]
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}
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}
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return dp[target-1]
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}
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```
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### **...**
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class Solution {
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public:
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int findTargetSumWays(vector<int>& nums, int target) {
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int s = 0;
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for (int x : nums) s += x;
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if (s - target < 0 || (s - target) % 2 != 0) return 0;
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target = (s - target) / 2 + 1;
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vector<int> dp(target);
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dp[0] = 1;
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for (int i = 1; i < nums.size() + 1; ++i)
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{
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for (int j = target - 1; j >= nums[i - 1]; --j)
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{
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dp[j] += dp[j - nums[i - 1]];
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}
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}
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return dp[target - 1];
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}
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};
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func findTargetSumWays(nums []int, target int) int {
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s := 0
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for _, x := range nums {
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s += x
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}
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if s-target < 0 || (s-target)%2 != 0 {
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return 0
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}
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target = (s-target)/2 + 1
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dp := make([]int, target)
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dp[0] = 1
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for i := 1; i < len(nums)+1; i++ {
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for j := target - 1; j >= nums[i-1]; j-- {
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dp[j] += dp[j-nums[i-1]]
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}
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}
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return dp[target-1]
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}
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class Solution {
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public int findTargetSumWays(int[] nums, int target) {
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int s = 0;
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for (int x : nums) {
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s += x;
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}
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if (s - target < 0 || (s - target) % 2 != 0) {
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return 0;
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}
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target = (s - target) / 2 + 1;
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int[] dp = new int[target];
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dp[0] = 1;
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for (int i = 1; i < nums.length + 1; ++i) {
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for (int j = target - 1; j >= nums[i - 1]; --j) {
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dp[j] += dp[j - nums[i - 1]];
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}
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}
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return dp[target - 1];
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}
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}
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class Solution:
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def findTargetSumWays(self, nums: List[int], target: int) -> int:
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s = sum(nums)
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if s - target < 0 or (s - target) % 2 != 0:
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return 0
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target = (s - target) // 2 + 1
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n = len(nums) + 1
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dp = [0] * target
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dp[0] = 1
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for i in range(1, n):
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for j in range(target - 1, nums[i - 1] - 1, -1):
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dp[j] += dp[j - nums[i - 1]]
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return dp[-1]

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