3.1 Java实现 publicclassSolution{publicbooleancanPartition(int[] nums){// 数组求和intsum=Arrays.stream(nums).sum();// 场景1:和为奇数不能均分if(sum %2==1) {returnfalse; }inttarget=sum /2;intn=nums.length;boolean[][] dp =newboolean[n +1][target +1]; dp[0][0] =true;for(inti...
int sum = Arrays.stream(nums).sum(); // 场景1:和为奇数不能均分 if (sum % 2 == 1) { return false; } int target = sum / 2; int n = nums.length; boolean[][] dp = new boolean[n + 1][target + 1]; dp[0][0] = true; for (int i = 1; i <= n; i++) { for (...
return false; int sum = 0, target = 0; for (int i = 0; i < len; i++) { sum += nums[i]; } if (sum & 1) //如果sum是奇数,直接return false { return false; } else //如果是偶数,说明可能为true 可能为false,如果能凑出若干整数使它们和为sum/2,说明为true { target = sum / ...
Explanation: The array cannot be partitioned into equal sum subsets. 解法一: 一上来首先想到的并不是动态规划,而是广度优先搜索。首先观察一下题目,数组能被对半分的前提是数组元素和必须是偶数。那么本题就转化成了之前做过的一个问题,即LeetCode 40. Combination Sum II —— 组合总和 II,寻找target为数组...
解:1.nums和一定是被2整除。2.定义一个一维的dp数组,其中dp[i]表示数字i是否是原数组的任意个子集合之和,那么我们我们最后只需要返回dp[target]就行了。我...
416. Partition Equal Subset Sum # 题目 # 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: Each of the array element will not
int sum=accumulate(nums.begin(),nums.end(),0); if(sum&1)//奇数 return false; int n=nums.size(); int target=sum>>1; vector<vector<bool>> dp(n,vector<bool>(target+1)); if(nums[0]<=target){ dp[0][nums[0]]=true;
Explanation: The array cannot be partitioned into equal sum subsets. 描述 给定一个只包含正整数的非空数组。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。 注意: 每个数组中的元素不会超过 100 数组的大小不会超过 200 示例1:
A set of practice note, solution, complexity analysis and test bench to leetcode problem set - leetcode/Subset_II_by_backtracking.drawio at b58bcceb0ea27d0756ad72fb6a64b3b547fae221 · brianchiang-tw/leetcode
Explanation: The array cannot be partitioned into equal sum subsets. Constraints: 1 <= nums.length <= 200 1 <= nums[i] <= 100 题目描述: 给定一个只包含正整数的非空数组。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。