LEETCODE之Combination Sum系列

Combination Sum

https://leetcode.com/problems/combination-sum/

题目中数值并不重复，而且允许使用多次，很简单的回溯算法，递归求解，注意先把数组排序一下，代码如下

class Solution {public:    vector<vector<int> > combinationSum(vector<int>& candidates, int target) {    vector<vector<int> > ivv;    vector<int> tmp;    sort(candidates.begin(),candidates.end());    combinationSum2(ivv,candidates,tmp,target,0);    return ivv;}void combinationSum2(vector<vector<int> >& ivv,vector<int>& candidates, vector<int>& tmp,int target,int j) {    if(target<0)return ;    if(target==0){ivv.push_back(tmp);return ;}    for(int i=j;i<candidates.size();++i){        if(candidates[i]<=target){            tmp.push_back(candidates[i]);            combinationSum2(ivv,candidates,tmp,target-candidates[i],i);            tmp.pop_back();        }    }}};

Combination Sum2

https://leetcode.com/problems/combination-sum-ii/

题目中有数值的重复而只允许使用一次，因此需要排除重复数值，和上一题的思路一样，关键点在与排除重复的操作应该在递归之后，代码如下

class Solution {public:    vector<vector<int> > combinationSum2(vector<int>& candidates,int target){vector<vector<int> > ret;vector<int> tmp;sort(candidates.begin(),candidates.end());helper(ret,tmp,candidates,target,0);return ret;    }    void helper(vector<vector<int> >&ret,vector<int>& tmp,vector<int>& nums,int k,int j){if(k<0) return ;if(k==0){ret.push_back(tmp);return;}for(int i=j;i<nums.size();++i){    if(nums[i]<=k){tmp.push_back(nums[i]);helper(ret,tmp,nums,k-nums[i],i+1);while(i<nums.size()&&nums[i]==nums[i+1]) {                      i++;                  }tmp.pop_back();    }}    }};

Combination Sum3

https://leetcode.com/problems/combination-sum-iii/

题目给的数组为[1-9]，区别于之前的题是增加了个数的限制，即k个不同的数构成n，求组合。思路一样，代码如下

class Solution {public:    vector<vector<int>> combinationSum3(int k, int n) {    vector<vector<int>> result;    vector<int> path;    backtrack(result, path, 1, k, n);    return result;}void backtrack(vector<vector<int>> &result, vector<int> &path, int start, int k, int target){    if(target==0&&k==0){        result.push_back(path);        return;    }    for(int i=start;i<=10-k&&i<=target;i++){        path.push_back(i);        backtrack(result,path,i+1,k-1,target-i);        path.pop_back();    }}//回溯，学习};

Combination Sum4

https://leetcode.com/problems/combination-sum-iv/

题目给的提示是动态规划，公式dp[i]=Σdp[i-nums[k]]  0<=k<=nums.size()

代码如下

class Solution {public:    int combinationSum4(vector<int>& nums, int target) {        vector<int> dp(target + 1);        dp[0] = 1;        sort(nums.begin(), nums.end());        for (int i = 1; i <= target; ++i) {            for (auto a : nums) {                if (i < a) break;                dp[i] += dp[i - a];            }        }        return dp.back();    }};