diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index c9f88892..32931e6b 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -293,19 +293,84 @@ class Solution { ``` ### Python +回溯版 +```python +class Solution: + + + def backtracking(self, candidates, target, total, startIndex, path, result): + if total == target: + result.append(path[:]) # 将当前路径的副本添加到结果中 + # 如果 sum + candidates[i] > target,则停止遍历 + for i in range(startIndex, len(candidates)): + if total + candidates[i] > target: + break + total += candidates[i] + path.append(candidates[i]) + self.backtracking(candidates, target, total, i + 1, path, result) + total -= candidates[i] + path.pop() + + def findTargetSumWays(self, nums: List[int], target: int) -> int: + total = sum(nums) + if target > total: + return 0 # 此时没有方案 + if (target + total) % 2 != 0: + 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 + bagSize = (target + total) // 2 # 转化为组合总和问题,bagSize就是目标和 + + # 以下是回溯法代码 + result = [] + nums.sort() # 需要对nums进行排序 + self.backtracking(nums, bagSize, 0, 0, [], result) + return len(result) + +``` +二维DP ```python class Solution: def findTargetSumWays(self, nums: List[int], target: int) -> int: - sumValue = sum(nums) - #注意边界条件为 target>sumValue or target<-sumValue or (sumValue + target) % 2 == 1 - if abs(target) > sumValue or (sumValue + target) % 2 == 1: return 0 - bagSize = (sumValue + target) // 2 - dp = [0] * (bagSize + 1) - dp[0] = 1 - for i in range(len(nums)): - for j in range(bagSize, nums[i] - 1, -1): - dp[j] += dp[j - nums[i]] - return dp[bagSize] + total_sum = sum(nums) # 计算nums的总和 + if abs(target) > total_sum: + return 0 # 此时没有方案 + if (target + total_sum) % 2 == 1: + return 0 # 此时没有方案 + target_sum = (target + total_sum) // 2 # 目标和 + + # 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 + dp = [[0] * (target_sum + 1) for _ in range(len(nums) + 1)] + + # 初始化状态 + dp[0][0] = 1 + + # 动态规划过程 + for i in range(1, len(nums) + 1): + for j in range(target_sum + 1): + dp[i][j] = dp[i - 1][j] # 不选取当前元素 + if j >= nums[i - 1]: + dp[i][j] += dp[i - 1][j - nums[i - 1]] # 选取当前元素 + + return dp[len(nums)][target_sum] # 返回达到目标和的方案数 + + +``` +一维DP +```python +class Solution: + def findTargetSumWays(self, nums: List[int], target: int) -> int: + total_sum = sum(nums) # 计算nums的总和 + if abs(target) > total_sum: + return 0 # 此时没有方案 + if (target + total_sum) % 2 == 1: + return 0 # 此时没有方案 + target_sum = (target + total_sum) // 2 # 目标和 + dp = [0] * (target_sum + 1) # 创建动态规划数组,初始化为0 + dp[0] = 1 # 当目标和为0时,只有一种方案,即什么都不选 + for num in nums: + for j in range(target_sum, num - 1, -1): + dp[j] += dp[j - num] # 状态转移方程,累加不同选择方式的数量 + return dp[target_sum] # 返回达到目标和的方案数 + ``` ### Go