Subsets and Subsets II (回溯，DFS，组合问题)

【Subsets and Subsets II (回溯，DFS，组合问题)】的更多相关文章

Subsets and Subsets II (回溯，DFS，组合问题)

Given a set of distinct integers, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example,If S = [1,2,3], a solution is: [ [3], [1], [2], [1,2,3], [1…

[LeetCode] Subsets I (78) & II (90) 解题思路，即全组合算法

78. Subsets Given a set of distinct integers, nums, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example,If nums = [1,2,3], a solution is: [ [3], [1]…

LeetCode Subsets II （DFS）

题意: 给一个集合,有n个可能相同的元素,求出所有的子集(包括空集,但是不能重复). 思路: 看这个就差不多了.LEETCODE SUBSETS (DFS) class Solution { public: vector<vector<int>> subsets(vector<int>& nums) { sort(nums.begin(),nums.end()); DFS(,nums,tmp); ans.push_back(vector<int>()…

78. Subsets 90. Subsets II

1. Given a set of distinct integers, nums, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example,If nums = [1,2,3], a solution is: [ [3], [1], [2], [1…

Different Subsets For All Tuples CodeForces - 660E (组合计数)

大意: 定义$f(a)$表示序列$a$本质不同子序列个数. 给定$n,m$, 求所有长$n$元素范围$[1,m]$的序列的$f$值之和. 显然长度相同的子序列贡献是相同的. 不考虑空串, 假设长$x$, 枚举第一次出现位置, 可以得到贡献为$\sum\limits_{i=x}^n\binom{i-1}{x-1}(m-1)^{i-x}m^{n-i}$ 总的答案就为$\sum\limits_{x=1}^n m^x \sum\limits_{i=x}^n\binom{i-1}{x-1}(m-1)^{i…

Path Sum II 总结DFS

https://oj.leetcode.com/problems/path-sum-ii/ Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return…

LeetCode Combination Sum II （DFS）

题意: 在集合candidates中选出任意多个元素,使得他们的和为target,返回所有的组合,以升序排列. 思路: 难点在于如何去重,比如集合{1,1,2},target=3,那么只有一个组合就是1+2=3,而不是两个. DFS解决,每次考虑candidates[i]取还是不取,但是这样子还是会产生重复,这里只需要一个技巧就可以使得没有重复出现.如果当前元素已经被考虑过取了,那么在考虑不取的时候,i后面的与candidates[i]相同的都要跳过.观察一下可以发现,如果有一串相同的数字出现的…