Leetcode 416. Partition Equal Subset Sum 回溯法 壓縮的一維動態規劃 位運算

1 回溯法

class Solution {
public:
    bool backtrack(vector<int>& nums, int start, int target) {
        if (target <= 0) return target == 0;
        for (int i = start; i < nums.size(); i++) 
            if (backtrack(nums, i + 1, target - nums[i])) return true;
        return false;
    }
    
    bool canPartition(vector<int>& nums) {
        int sum = accumulate(nums.begin(), nums.end(), 0);
        return !(sum & 1) && backtrack(nums, 0, sum >> 1);
    }
};

2 動態規劃

class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if(sum&1) return false;
        vector<int> dp(sum+1,0);
        dp[0]=1;
        for(auto num: nums){
            for(int i=sum; i>=0; i--){
                if(dp[i]) dp[i+num]=1;
            }
            if(dp[sum>>1]) return true;
        }
        return false;
    }
};

3 位運算

bool canPartition(vector<int>& nums) {
    bitset<5001> bits(1);
    int sum = accumulate(nums.begin(), nums.end(), 0);
    for (auto n : nums) bits |= bits << n;
    return !(sum & 1) && bits[sum >> 1];
}