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Update 0494.目标和.md
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@ -293,19 +293,84 @@ class Solution {
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```
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### Python
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回溯版
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```python
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class Solution:
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def backtracking(self, candidates, target, total, startIndex, path, result):
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if total == target:
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result.append(path[:]) # 将当前路径的副本添加到结果中
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# 如果 sum + candidates[i] > target,则停止遍历
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for i in range(startIndex, len(candidates)):
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if total + candidates[i] > target:
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break
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total += candidates[i]
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path.append(candidates[i])
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self.backtracking(candidates, target, total, i + 1, path, result)
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total -= candidates[i]
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path.pop()
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def findTargetSumWays(self, nums: List[int], target: int) -> int:
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total = sum(nums)
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if target > total:
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return 0 # 此时没有方案
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if (target + total) % 2 != 0:
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return 0 # 此时没有方案,两个整数相加时要注意数值溢出的问题
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bagSize = (target + total) // 2 # 转化为组合总和问题,bagSize就是目标和
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# 以下是回溯法代码
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result = []
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nums.sort() # 需要对nums进行排序
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self.backtracking(nums, bagSize, 0, 0, [], result)
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return len(result)
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```
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二维DP
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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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sumValue = sum(nums)
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#注意边界条件为 target>sumValue or target<-sumValue or (sumValue + target) % 2 == 1
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if abs(target) > sumValue or (sumValue + target) % 2 == 1: return 0
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bagSize = (sumValue + target) // 2
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dp = [0] * (bagSize + 1)
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dp[0] = 1
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for i in range(len(nums)):
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for j in range(bagSize, nums[i] - 1, -1):
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dp[j] += dp[j - nums[i]]
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return dp[bagSize]
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total_sum = sum(nums) # 计算nums的总和
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if abs(target) > total_sum:
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return 0 # 此时没有方案
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if (target + total_sum) % 2 == 1:
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return 0 # 此时没有方案
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target_sum = (target + total_sum) // 2 # 目标和
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# 创建二维动态规划数组,行表示选取的元素数量,列表示累加和
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dp = [[0] * (target_sum + 1) for _ in range(len(nums) + 1)]
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# 初始化状态
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dp[0][0] = 1
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# 动态规划过程
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for i in range(1, len(nums) + 1):
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for j in range(target_sum + 1):
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dp[i][j] = dp[i - 1][j] # 不选取当前元素
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if j >= nums[i - 1]:
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dp[i][j] += dp[i - 1][j - nums[i - 1]] # 选取当前元素
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return dp[len(nums)][target_sum] # 返回达到目标和的方案数
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```
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一维DP
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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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total_sum = sum(nums) # 计算nums的总和
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if abs(target) > total_sum:
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return 0 # 此时没有方案
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if (target + total_sum) % 2 == 1:
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return 0 # 此时没有方案
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target_sum = (target + total_sum) // 2 # 目标和
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dp = [0] * (target_sum + 1) # 创建动态规划数组,初始化为0
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dp[0] = 1 # 当目标和为0时,只有一种方案,即什么都不选
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for num in nums:
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for j in range(target_sum, num - 1, -1):
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dp[j] += dp[j - num] # 状态转移方程,累加不同选择方式的数量
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return dp[target_sum] # 返回达到目标和的方案数
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```
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### Go
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