跳转至

剑指 Offer II 101. 分割等和子串

题目描述

给定一个非空的正整数数组 nums ，请判断能否将这些数字分成元素和相等的两部分。

 

示例 1：

输入：nums = [1,5,11,5]
输出：true
解释：nums 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：nums 不可以分为和相等的两部分

 

提示：

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

 

注意：本题与主站 416 题相同： https://leetcode.cn/problems/partition-equal-subset-sum/

解法

方法一

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        s = sum(nums)
        if s % 2 != 0:
            return False

        m, n = len(nums), (s >> 1) + 1
        dp = [[False] * n for _ in range(m)]
        for i in range(m):
            dp[i][0] = True
        if nums[0] < n:
            dp[0][nums[0]] = True

        for i in range(1, m):
            for j in range(n):
                dp[i][j] = dp[i - 1][j]
                if not dp[i][j] and nums[i] <= j:
                    dp[i][j] = dp[i - 1][j - nums[i]]
        return dp[-1][-1]

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
    public boolean canPartition(int[] nums) {
        int s = 0;
        for (int x : nums) {
            s += x;
        }
        if (s % 2 != 0) {
            return false;
        }
        int m = nums.length, n = (s >> 1) + 1;
        boolean[] dp = new boolean[n];
        dp[0] = true;
        if (nums[0] < n) {
            dp[nums[0]] = true;
        }
        for (int i = 1; i < m; ++i) {
            for (int j = n - 1; j >= nums[i]; --j) {
                dp[j] = dp[j] || dp[j - nums[i]];
            }
        }
        return dp[n - 1];
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int s = 0;
        for (int x : nums) s += x;
        if (s % 2 != 0) return false;
        int m = nums.size(), n = (s >> 1) + 1;
        vector<bool> dp(n);
        dp[0] = true;
        if (nums[0] < n) dp[nums[0]] = true;
        for (int i = 1; i < m; ++i) {
            for (int j = n - 1; j >= nums[i]; --j) {
                dp[j] = dp[j] || dp[j - nums[i]];
            }
        }
        return dp[n - 1];
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
func canPartition(nums []int) bool {
    s := 0
    for _, x := range nums {
        s += x
    }
    if s%2 != 0 {
        return false
    }
    m, n := len(nums), (s>>1)+1
    dp := make([]bool, n)
    dp[0] = true
    if nums[0] < n {
        dp[nums[0]] = true
    }
    for i := 1; i < m; i++ {
        for j := n - 1; j >= nums[i]; j-- {
            dp[j] = dp[j] || dp[j-nums[i]]
        }
    }
    return dp[n-1]
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution {
    func canPartition(_ nums: [Int]) -> Bool {
        let s = nums.reduce(0, +)
        if s % 2 != 0 { return false }
        let target = s / 2
        var dp = Array(repeating: false, count: target + 1)
        dp[0] = true

        for num in nums {
            for j in stride(from: target, through: num, by: -1) {
                dp[j] = dp[j] || dp[j - num]
            }
        }

        return dp[target]
    }
}

方法二

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        s = sum(nums)
        if s % 2 != 0:
            return False

        m, n = len(nums), (s >> 1) + 1
        dp = [False] * n
        dp[0] = True
        if nums[0] < n:
            dp[nums[0]] = True

        for i in range(1, m):
            for j in range(n - 1, nums[i] - 1, -1):
                dp[j] = dp[j] or dp[j - nums[i]]
        return dp[-1]

方法三

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        s = sum(nums)
        if s % 2 != 0:
            return False
        target = s >> 1

        @cache
        def dfs(i, s):
            nonlocal target
            if s > target or i >= len(nums):
                return False
            if s == target:
                return True
            return dfs(i + 1, s) or dfs(i + 1, s + nums[i])

        return dfs(0, 0)

评论