可以用来解决的问题有: Leetcode 78. Subsets , Leetcode 90. Subsets II, Leetcode 46. Permutations, Leetcode 47. Permutations II(contains duplicates), Leetcode 39. Combination Sum, Leetcode 40. Combination Sum II, Leetcode 131. Palindrome Partitioning. 好文要顶 关注我 收藏该文 微信分享 Joh...
#include <algorithm> using namespace std; class Solution{ public: vector<vector<int> > combinationSum2(vector<int> &candidates, int target){ vector<vector<int> > ans; vector<int> vecCur; sort(candidates.begin(), candidates.end()); combinationSum2Backtracking(candidates, 0, vecCur, target, ...
Then we will choose any of the value from starting and start our backtracking algorithm according to that and find the subsets with equal sum, {85,21,23,70,85} {28,59,31,93,73} {29,51,25,97,82} C++ Implementation #include<bits/stdc++.h>usingnamespacestd;boolis_partiable(int*arr,...
Permutations Subsets是「组合」,Permutations是排列。 那么这题也容易想到,跟subsets相比,把进入下一层dfs的i+1去掉就好了(同时要加上terminator哦)。但是这样一来对于[1,2,3]这样的nums,第一个解会变成[1,1,1]呀。怎么办,增加一个数组来保存是否使用过(模拟HashMap),有点像图的遍历了。或者直接利用ArrayList...