http://bgmeow.xyz/2017/01/29/LeetCode-494/ 1defsubsetSum(nums, target):2dp = [0]*(target+1)3dp[0] = 14fornuminnums:5forjinrange(target, num-1, -1):6dp[j] += dp[j-num]7#print ("dp: ", target, j, dp[j])8returndp[target]910sumN =sum(nums)1112ifsumN < Sor(S+su...
int sum = accumulate(nums.begin(),nums.end(),0); if (sum < target || (sum + target) % 2 == 1) return 0; int count = subSetSum(nums, (sum + target) / 2); return count; } int subSetSum(vector<int> a, int sum) { vector<int> dp(sum + 1, 0); dp[0] = 1; for (...
dfs(nums, s, k+1, sum-nums[k]); } } DP: publicclassSolution {publicintfindTargetSumWays(int[] nums,ints) {intsum = 0;for(intn : nums) sum+=n;//两种情况找不到结果,找得到的话就用subsetSum去找,证书和是(s + sum) >>> 1,也就是除以2returnsum < s || (s + sum) % 2 >...
The sum of elements in the given array will not exceed 1000. Your output answer is guaranteed to be fitted in a 32-bit integer. Idea: DP Solution 1: DP Time complexity: O(n*sum) Space complexity: O(n*sum) C++ C++ SC O(n) ...
0?0:subsetSum(nums,(s+sum)>>>1);}publicintsubsetSum(int[]nums,int s){int[]dp=newint[s+1];dp[0]=1;// 初始记录0的位置为1for(int n:nums)// 对每个元素,看看他现有能和别的元素相加得到哪些位置的数for(int i=s;i>=n;i--)dp[i]+=dp[i-n];returndp[s];}}...
等号两边都加上sum(N) + sum(P),于是可以得到结果2 * sum(P) = target + sum(nums),那么这道题就转换成了,能否在数组中找到这样一个集合,和等于(target + sum(nums)) / 2。那么这题就转化为了第 416 题了。dp[i]中存储的是能使和为i的方法个数。
intS){intsum=accumulate(nums.begin(),nums.end(),0);returnsum<S||(S+sum)%2==1?0:subsetSum((S+sum)/2,nums);}intsubsetSum(ints,vector<int>&nums){intn=nums.size();vector<int>dp(s+1,0);dp[0]=1;for(intn:nums)for(inti=s;i>=n;i--)dp[i]+=dp[i-n];returndp[s];}...
publicintfindTargetSumWays(int[]nums,ints){intsum=0;for(intn:nums)sum+=n;returnsum0?0:subsetSum(nums,(s+sum)>>>1);}publicintsubsetSum(int[]nums,ints){int[]dp=newint[s+1];dp[0]=1;for(intn:nums)for(inti=s;i>=n;i--)dp[i]+=dp[i-n];returndp[s];} 常规解法:(遍历)...
and do not take all possible interactions into account! Thus, if you are predicting interactions for a subregion only, the results are based on the respective subset of interactions!This model is used by the IntaRNAens personality.back to overview ...
sum = (sum + S) /2;returnsubsetsum(nums, sum); }intsubsetsum(vector<int>& nums,intsum){intdp[sum +1] = {0}; dp[0] =1;for(intn : nums) {for(intj = sum; j >= n; j--) dp[j] += dp[j - n]; }returndp[sum]; ...