Question

416. 分割等和子集

English Version

题目描述

给你一个 只包含正整数 的 非空 数组 nums 。请你判断是否可以将这个数组分割成两个子集，使得两个子集的元素和相等。

 

示例 1：

输入：nums = [1,5,11,5]
输出：true
解释：数组可以分割成 [1, 5, 5] 和 [11] 。

示例 2：

输入：nums = [1,2,3,5]
输出：false
解释：数组不能分割成两个元素和相等的子集。

 

提示：

• 1 <= nums.length <= 200
• 1 <= nums[i] <= 100

解法

方法一：动态规划

我们先计算出数组的总和 $s$，如果总和是奇数，那么一定不能分割成两个和相等的子集，直接返回 $false$。如果总和是偶数，我们记目标子集的和为 $m = \frac{s}{2}$，那么问题就转化成了：是否存在一个子集，使得其元素的和为 $m$。

我们定义 $f[i][j]$ 表示前 $i$ 个数中选取若干个数，使得其元素的和恰好为 $j$。初始时 $f[0][0] = true$，其余 $f[i][j] = false$。答案为 $f[n][m]$。

考虑 $f[i][j]$，如果我们选取了第 $i$ 个数 $x$，那么 $f[i][j] = f[i - 1][j - x]$；如果我们没有选取第 $i$ 个数 $x$，那么 $f[i][j] = f[i - 1][j]$。因此状态转移方程为：

$$ f[i][j] = f[i - 1][j] \text{ or } f[i - 1][j - x] \text{ if } j \geq x $$

最终答案为 $f[n][m]$。

注意到 $f[i][j]$ 只与 $f[i - 1][\cdot]$ 有关，因此我们可以将二维数组压缩成一维数组。

时间复杂度 $O(n \times m)$，空间复杂度 $O(m)$。其中 $n$ 是数组的长度，而 $m$ 是数组的总和的一半。

class Solution:
  def canPartition(self, nums: List[int]) -> bool:
    m, mod = divmod(sum(nums), 2)
    if mod:
      return False
    n = len(nums)
    f = [[False] * (m + 1) for _ in range(n + 1)]
    f[0][0] = True
    for i, x in enumerate(nums, 1):
      for j in range(m + 1):
        f[i][j] = f[i - 1][j] or (j >= x and f[i - 1][j - x])
    return f[n][m]

class Solution {
  public boolean canPartition(int[] nums) {
    // int s = Arrays.stream(nums).sum();
    int s = 0;
    for (int x : nums) {
      s += x;
    }
    if (s % 2 == 1) {
      return false;
    }
    int n = nums.length;
    int m = s >> 1;
    boolean[][] f = new boolean[n + 1][m + 1];
    f[0][0] = true;
    for (int i = 1; i <= n; ++i) {
      int x = nums[i - 1];
      for (int j = 0; j <= m; ++j) {
        f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
      }
    }
    return f[n][m];
  }
}

class Solution {
public:
  bool canPartition(vector<int>& nums) {
    int s = accumulate(nums.begin(), nums.end(), 0);
    if (s % 2 == 1) {
      return false;
    }
    int n = nums.size();
    int m = s >> 1;
    bool f[n + 1][m + 1];
    memset(f, false, sizeof(f));
    f[0][0] = true;
    for (int i = 1; i <= n; ++i) {
      int x = nums[i - 1];
      for (int j = 0; j <= m; ++j) {
        f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
      }
    }
    return f[n][m];
  }
};

func canPartition(nums []int) bool {
  s := 0
  for _, x := range nums {
    s += x
  }
  if s%2 == 1 {
    return false
  }
  n, m := len(nums), s>>1
  f := make([][]bool, n+1)
  for i := range f {
    f[i] = make([]bool, m+1)
  }
  f[0][0] = true
  for i := 1; i <= n; i++ {
    x := nums[i-1]
    for j := 0; j <= m; j++ {
      f[i][j] = f[i-1][j] || (j >= x && f[i-1][j-x])
    }
  }
  return f[n][m]
}

function canPartition(nums: number[]): boolean {
  const s = nums.reduce((a, b) => a + b, 0);
  if (s % 2 === 1) {
    return false;
  }
  const n = nums.length;
  const m = s >> 1;
  const f: boolean[][] = Array(n + 1)
    .fill(0)
    .map(() => Array(m + 1).fill(false));
  f[0][0] = true;
  for (let i = 1; i <= n; ++i) {
    const x = nums[i - 1];
    for (let j = 0; j <= m; ++j) {
      f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
    }
  }
  return f[n][m];
}

impl Solution {
  #[allow(dead_code)]
  pub fn can_partition(nums: Vec<i32>) -> bool {
    let mut sum = 0;
    for e in &nums {
      sum += *e;
    }

    if sum % 2 != 0 {
      return false;
    }

    let n = nums.len();
    let m = (sum / 2) as usize;
    let mut dp: Vec<Vec<bool>> = vec![vec![false; m + 1]; n + 1];

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

    // Begin the actual dp process
    for i in 1..=n {
      for j in 0..=m {
        dp[i][j] = if (nums[i - 1] as usize) > j {
          dp[i - 1][j]
        } else {
          dp[i - 1][j] || dp[i - 1][j - (nums[i - 1] as usize)]
        };
      }
    }

    dp[n][m]
  }
}

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var canPartition = function (nums) {
  const s = nums.reduce((a, b) => a + b, 0);
  if (s % 2 === 1) {
    return false;
  }
  const n = nums.length;
  const m = s >> 1;
  const f = Array(n + 1)
    .fill(0)
    .map(() => Array(m + 1).fill(false));
  f[0][0] = true;
  for (let i = 1; i <= n; ++i) {
    const x = nums[i - 1];
    for (let j = 0; j <= m; ++j) {
      f[i][j] = f[i - 1][j] || (j >= x && f[i - 1][j - x]);
    }
  }
  return f[n][m];
};

方法二

class Solution:
  def canPartition(self, nums: List[int]) -> bool:
    m, mod = divmod(sum(nums), 2)
    if mod:
      return False
    f = [True] + [False] * m
    for x in nums:
      for j in range(m, x - 1, -1):
        f[j] = f[j] or f[j - x]
    return f[m]

class Solution {
  public boolean canPartition(int[] nums) {
    // int s = Arrays.stream(nums).sum();
    int s = 0;
    for (int x : nums) {
      s += x;
    }
    if (s % 2 == 1) {
      return false;
    }
    int m = s >> 1;
    boolean[] f = new boolean[m + 1];
    f[0] = true;
    for (int x : nums) {
      for (int j = m; j >= x; --j) {
        f[j] |= f[j - x];
      }
    }
    return f[m];
  }
}

class Solution {
public:
  bool canPartition(vector<int>& nums) {
    int s = accumulate(nums.begin(), nums.end(), 0);
    if (s % 2 == 1) {
      return false;
    }
    int m = s >> 1;
    bool f[m + 1];
    memset(f, false, sizeof(f));
    f[0] = true;
    for (int& x : nums) {
      for (int j = m; j >= x; --j) {
        f[j] |= f[j - x];
      }
    }
    return f[m];
  }
};

func canPartition(nums []int) bool {
  s := 0
  for _, x := range nums {
    s += x
  }
  if s%2 == 1 {
    return false
  }
  m := s >> 1
  f := make([]bool, m+1)
  f[0] = true
  for _, x := range nums {
    for j := m; j >= x; j-- {
      f[j] = f[j] || f[j-x]
    }
  }
  return f[m]
}

function canPartition(nums: number[]): boolean {
  const s = nums.reduce((a, b) => a + b, 0);
  if (s % 2 === 1) {
    return false;
  }
  const m = s >> 1;
  const f: boolean[] = Array(m + 1).fill(false);
  f[0] = true;
  for (const x of nums) {
    for (let j = m; j >= x; --j) {
      f[j] = f[j] || f[j - x];
    }
  }
  return f[m];
}

impl Solution {
  #[allow(dead_code)]
  pub fn can_partition(nums: Vec<i32>) -> bool {
    let mut sum = 0;
    for e in &nums {
      sum += *e;
    }

    if sum % 2 != 0 {
      return false;
    }

    let m = (sum >> 1) as usize;

    // Here dp[i] means if it can be sum up to `i` for all the number we've traversed through so far
    // Which is actually compressing the 2-D dp vector to 1-D
    let mut dp: Vec<bool> = vec![false; m + 1];

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

    // Begin the actual dp process
    for e in &nums {
      // For every num in nums vector
      for i in (*e as usize..=m).rev() {
        // Update the current status
        dp[i] |= dp[i - (*e as usize)];
      }
    }

    dp[m]
  }
}

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var canPartition = function (nums) {
  const s = nums.reduce((a, b) => a + b, 0);
  if (s % 2 === 1) {
    return false;
  }
  const m = s >> 1;
  const f = Array(m + 1).fill(false);
  f[0] = true;
  for (const x of nums) {
    for (let j = m; j >= x; --j) {
      f[j] = f[j] || f[j - x];
    }
  }
  return f[m];
};