416. Partition Equal Subset Sum

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Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.

Note:

  1. Each of the array element will not exceed 100.
  2. The array size will not exceed 200.

Example 1:

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Input: [1, 5, 11, 5]

Output: true

Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:

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Input: [1, 2, 3, 5]

Output: false

Explanation: The array cannot be partitioned into equal sum subsets.

解法1：动态规划

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

时间复杂度：与nums.size()和target都相关。

解法2

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class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int sum = 0;
        bool res = false;
        if(nums.size() < 2){
            return false;
        }
        for(int i=0;i<nums.size();++i){
            sum += nums[i];
        }
        if(sum & 1){
            return false;
        }
        sort(nums.begin(), nums.end(), less<int>());
        for(int i=0;i<nums.size();++i){
            int tmp = sum >> 1;
            int end = nums.size() - 1;
            tmp = tmp - nums[i];
            if(tmp == 0){
                res = true;
            }
            while(i < end){
                if(tmp - nums[end] > 0){
                    tmp = tmp - nums[end];
                    end--;
                }else if(tmp - nums[end] < 0){
                    end--;
                }else{
                    res = true;
                    end--;
                }
            }
        }
        return res;
    }
};

时间复杂度：仅仅与nums.size()相关