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=1; i <= ...
Can you solve this real interview question? Partition Equal Subset Sum - Given an integer array nums, return true if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or false otherwise. Example 1
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 (...
https://discuss.leetcode.com/topic/46161/a-general-approach-to-backtracking-questions-in-java-subsets-permutations-combination-sum-palindrome-partitioning/2 里面比较难想的部分(对于我这种只捡easy模式的题目做的算法小白)是循环里面的递归,每次退栈的时候,会从cur中remove一个元素出来,然后i要加1,继续循环!!
思路1:动态规划 class Solution { public: bool canPartition(vector<int>& nums) { 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)); ...
如果数组长度为N,目标sum(即总和的一半)是M,由于全部是正整数,那么在递推过程中涉及到的和只可能是0到M,于是可以用一个 N x (M+1) 的表格tab记录结果。其中tab[i][j]表示在第0至i个数中,是否存在和为j的子集。时间复杂度为O(MN),因为每次递推只需要用到上一行的结果,则空间复杂度可以优化到O(M)...
[leetcode]416. Partition Equal Subset Sum 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. 给定一个非空数组,是否能把数组划分为两个和相等的子集。 Not......
输入: [1, 2, 3, 5] 输出: false 解释: 数组无法被拆分成两个和相等的子数组. 想法 首先我们需要遍历一趟得到整个数组的和sum,从而求出每个子数组元素的加和(sum/2)。 如果sum是奇数的话,那一定是不满足条件的,可以直接返回false。如果sum是偶数,将sum除2获得我们要求的一个子数组元素的和。
来自专栏 · LeetCodeDescription 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 exceed 100.The array size will not exceed 200....
Leetcode每日一题:416.partition-equal-subset-sum(分割等和子集),思路:这题从动态规划的思想上来看很像0-1背包问题,后者需要小于等于背包容量的条件下价值最大化,这里则是刚好等于数组之和的一半;1°,要想满足条件,数组之和sum必须为偶数,并且目标值target=sum/