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发布于 2024-06-17 01:04:00 字数 8513 浏览 0 评论 0 收藏 0

494. 目标和

English Version

题目描述

给你一个非负整数数组 nums 和一个整数 target

向数组中的每个整数前添加 '+''-' ,然后串联起所有整数,可以构造一个 表达式

  • 例如,nums = [2, 1] ,可以在 2 之前添加 '+' ,在 1 之前添加 '-' ,然后串联起来得到表达式 "+2-1"

返回可以通过上述方法构造的、运算结果等于 target 的不同 表达式 的数目。

 

示例 1:

输入:nums = [1,1,1,1,1], target = 3
输出:5
解释:一共有 5 种方法让最终目标和为 3 。
-1 + 1 + 1 + 1 + 1 = 3
+1 - 1 + 1 + 1 + 1 = 3
+1 + 1 - 1 + 1 + 1 = 3
+1 + 1 + 1 - 1 + 1 = 3
+1 + 1 + 1 + 1 - 1 = 3

示例 2:

输入:nums = [1], target = 1
输出:1

 

提示:

  • 1 <= nums.length <= 20
  • 0 <= nums[i] <= 1000
  • 0 <= sum(nums[i]) <= 1000
  • -1000 <= target <= 1000

解法

方法一:动态规划

题目可以转换为 0-1 背包问题。

设整数数组总和为 s,添加 - 号的元素之和为 x,则添加 + 号的元素之和为 s - x,那么 s - x - x = target2x = s - target。左式成立需要满足 s - target 一定大于等于 0,并且能够被 2 整除。在此前提下,我们可以将问题抽象为: 从数组中选出若干个数,使得选出的元素之和为 x。显然这是一个 0-1 背包问题。

定义 dp[i][j] 表示从前 i 个数中选出若干个数,使得所选元素之和为 j 的所有方案数。

class Solution:
  def findTargetSumWays(self, nums: List[int], target: int) -> int:
    s = sum(nums)
    if s < target or (s - target) % 2 != 0:
      return 0
    m, n = len(nums), (s - target) // 2
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = 1
    for i in range(1, m + 1):
      for j in range(n + 1):
        dp[i][j] = dp[i - 1][j]
        if nums[i - 1] <= j:
          dp[i][j] += dp[i - 1][j - nums[i - 1]]
    return dp[-1][-1]
class Solution {
  public int findTargetSumWays(int[] nums, int target) {
    int s = 0;
    for (int v : nums) {
      s += v;
    }
    if (s < target || (s - target) % 2 != 0) {
      return 0;
    }
    int m = nums.length;
    int n = (s - target) / 2;
    int[][] dp = new int[m + 1][n + 1];
    dp[0][0] = 1;
    for (int i = 1; i <= m; ++i) {
      for (int j = 0; j <= n; ++j) {
        dp[i][j] = dp[i - 1][j];
        if (nums[i - 1] <= j) {
          dp[i][j] += dp[i - 1][j - nums[i - 1]];
        }
      }
    }
    return dp[m][n];
  }
}
class Solution {
public:
  int findTargetSumWays(vector<int>& nums, int target) {
    int s = accumulate(nums.begin(), nums.end(), 0);
    if (s < target || (s - target) % 2 != 0) return 0;
    int m = nums.size(), n = (s - target) / 2;
    vector<vector<int>> dp(m + 1, vector<int>(n + 1));
    dp[0][0] = 1;
    for (int i = 1; i <= m; ++i) {
      for (int j = 0; j <= n; ++j) {
        dp[i][j] += dp[i - 1][j];
        if (nums[i - 1] <= j) dp[i][j] += dp[i - 1][j - nums[i - 1]];
      }
    }
    return dp[m][n];
  }
};
func findTargetSumWays(nums []int, target int) int {
  s := 0
  for _, v := range nums {
    s += v
  }
  if s < target || (s-target)%2 != 0 {
    return 0
  }
  m, n := len(nums), (s-target)/2
  dp := make([][]int, m+1)
  for i := range dp {
    dp[i] = make([]int, n+1)
  }
  dp[0][0] = 1
  for i := 1; i <= m; i++ {
    for j := 0; j <= n; j++ {
      dp[i][j] = dp[i-1][j]
      if nums[i-1] <= j {
        dp[i][j] += dp[i-1][j-nums[i-1]]
      }
    }
  }
  return dp[m][n]
}
impl Solution {
  #[allow(dead_code)]
  pub fn find_target_sum_ways(nums: Vec<i32>, target: i32) -> i32 {
    let mut sum = 0;
    for e in &nums {
      sum += *e;
    }

    // -x + (sum - x) = target <-> -2 * x + sum = target <-> 2 * x = sum - target
    if sum < target || (sum - target) % 2 != 0 {
      // There is no way to get any expression in this case
      return 0;
    }
    let n = nums.len();
    let m = (sum - target) / 2;

    let mut dp: Vec<Vec<i32>> = vec![vec![0; m as usize + 1]; n + 1];

    // Initialize the dp vector
    dp[0][0] = 1;

    // Begin the actual dp phase
    for i in 1..=n {
      for j in 0..=m as usize {
        // nums[i - 1] is not included
        dp[i][j] = dp[i - 1][j];
        if nums[i - 1] <= (j as i32) {
          // nums[i - 1] is included
          dp[i][j] += dp[i - 1][j - (nums[i - 1] as usize)];
        }
      }
    }

    dp[n][m as usize]
  }
}
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var findTargetSumWays = function (nums, target) {
  let s = 0;
  for (let v of nums) {
    s += v;
  }
  if (s < target || (s - target) % 2 != 0) {
    return 0;
  }
  const m = nums.length;
  const n = (s - target) / 2;
  let dp = new Array(n + 1).fill(0);
  dp[0] = 1;
  for (let i = 1; i <= m; ++i) {
    for (let j = n; j >= nums[i - 1]; --j) {
      dp[j] += dp[j - nums[i - 1]];
    }
  }
  return dp[n];
};

方法二

class Solution:
  def findTargetSumWays(self, nums: List[int], target: int) -> int:
    s = sum(nums)
    if s < target or (s - target) % 2 != 0:
      return 0
    n = (s - target) // 2
    dp = [0] * (n + 1)
    dp[0] = 1
    for v in nums:
      for j in range(n, v - 1, -1):
        dp[j] += dp[j - v]
    return dp[-1]
class Solution {
  public int findTargetSumWays(int[] nums, int target) {
    int s = 0;
    for (int v : nums) {
      s += v;
    }
    if (s < target || (s - target) % 2 != 0) {
      return 0;
    }
    int n = (s - target) / 2;
    int[] dp = new int[n + 1];
    dp[0] = 1;
    for (int v : nums) {
      for (int j = n; j >= v; --j) {
        dp[j] += dp[j - v];
      }
    }
    return dp[n];
  }
}
class Solution {
public:
  int findTargetSumWays(vector<int>& nums, int target) {
    int s = accumulate(nums.begin(), nums.end(), 0);
    if (s < target || (s - target) % 2 != 0) return 0;
    int n = (s - target) / 2;
    vector<int> dp(n + 1);
    dp[0] = 1;
    for (int& v : nums)
      for (int j = n; j >= v; --j)
        dp[j] += dp[j - v];
    return dp[n];
  }
};
func findTargetSumWays(nums []int, target int) int {
  s := 0
  for _, v := range nums {
    s += v
  }
  if s < target || (s-target)%2 != 0 {
    return 0
  }
  n := (s - target) / 2
  dp := make([]int, n+1)
  dp[0] = 1
  for _, v := range nums {
    for j := n; j >= v; j-- {
      dp[j] += dp[j-v]
    }
  }
  return dp[n]
}
impl Solution {
  #[allow(dead_code)]
  pub fn find_target_sum_ways(nums: Vec<i32>, target: i32) -> i32 {
    let mut sum = 0;
    for e in &nums {
      sum += *e;
    }

    if sum < target || (sum - target) % 2 != 0 {
      // Just directly return
      return 0;
    }

    let n = ((sum - target) / 2) as usize;
    let mut dp: Vec<i32> = vec![0; n + 1];

    // Initialize the dp vector
    dp[0] = 1;

    // Begin the actual dp phase
    for e in &nums {
      for i in (*e as usize..=n).rev() {
        dp[i] += dp[i - (*e as usize)];
      }
    }

    dp[n]
  }
}

方法三

class Solution:
  def findTargetSumWays(self, nums: List[int], target: int) -> int:
    @cache
    def dfs(i, t):
      if i == n:
        if t == target:
          return 1
        return 0
      return dfs(i + 1, t + nums[i]) + dfs(i + 1, t - nums[i])

    ans, n = 0, len(nums)
    return dfs(0, 0)

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