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...
代码如下: classSolution(object):deffindTargetSumWays(self, nums, S):""":type nums: List[int] :type S: int :rtype: int"""Amount=sum(nums)ifS > AmountorS < -Amount:return0 MaxValue=max(nums) dp= [[0forxinxrange(2*(Amount+MaxValue)+1)]forxinnums] dp[0][nums[0]+ Amount] +...
/* * 描述:使用备忘录方法对整个树进行遍历,尽可能地减少重复子问题的出现 * 参数:nums是需要遍历的数组 * sum是当前计算出来的总和 * index是当前遍历的索引,递归迭代的条件 * target是要计算的目标 * note是记录递归过程的备忘录,是一个二维数组 * 返回:所有等于target的最终的总和 */ int calculate(vector...
int findTargetSumWays(vector<int>& nums, int S) { const int N = nums.size(); vector<unordered_map<int, int>> dp(N + 1); dp[0][0] = 1; for (int i = 0; i < N; i ++) { for (auto m : dp[i]) { dp[i + 1][m.first + nums[i]] += m.second; dp[i + 1][...
dp[i] += dp[i-num]; } } return dp[n]; } };本题应用动态编程思想解,这个解参考了discuss中一位大神的解,问题的转化十分巧妙。 具体的分析参考:https://discuss.leetcode.com/topic/76243/java-15-ms-c-3-ms-o-ns-iterative-dp-solution-using-subset-sum-with-explanation?page=1版权...
最后的和S也要映射到S+sum public int findTargetSumWays(int[] nums, int S) { if(nums == null || nums.length == 0) { return 0; } int sum = 0; for(int i = 0; i < nums.length; i++) { sum += nums[i]; } int[][] dp = new int[nums.length + 1][2*sum + 1]; ...
还有就是通过dp来做,解法如下:https://leetcode.com/problems/target-sum/discuss/97335/Short-Java-DP-Solution-with-Explanation this is a classic knapsack problem in knapsack, we decide whether we choose this element or not in this question, we decide whether we add this element or minus it ...
fill(dp.begin(), dp.end(), -1); return solve(nums, target,0); } int solve(vector<int>& nums, int target, int currSum){ if(currSum > target) return 0; if(currSum == target) return 1; if(dp[currSum] != -1) return dp[currSum]; ...
Target Sum 早上发那题居然被点名了zzz血崩 题目地址:https://leetcode.com/problems/range-sum-query-immutable/?tab=Description 题意是大概求数组内部分和达到指定和,典型dp问题 用了暴力遍历,代码如下 后来看了一下别人solution,用了一个非常巧妙的办法...
可以将 dp 优化为一维数组 dp[j] = ∑[max(0, j - k)], 每次遍历时需要将 dp[j] 置 0 时间复杂度为 O(nkt), 空间复杂度为 O(t) 代码: C++: class Solution{public:intnumRollsToTarget(intn,intk,inttarget){constintmod=1e9+7;vector<int>dp(target+1);dp.front()=1;for(inti=0;i<...