backTrack

leetcode中常见的backTrack类题目：combination、subsets、permutation、Palindrome Partitioning.

1、combination

<1>leetcode39. Combination Sum

题目描述：

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

void backTrack(vector<vector<int>>& ans, vector<int>& tmp, const vector<int>& nums, int remain, int start){
    if(remain < 0) return;
    else if(remain == 0) ans.push_back(tmp);
    else{
        for(int i = start; i < nums.size(); i++){
            tmp.push_back(nums[i]);
            backTrack(ans, tmp, nums, remain - nums[i], i);
            tmp.pop_back();
        }
    }
}

vector<vector<int>> combinationSum(vector<int>& nums, int target){
    vector<vector<int>> ans;
    vetcot<int> temp;
    sort(nums.begin(), nums.end());
    backTrack(ans, temp, nums, target, 0);
    return ans;
}

<2>leetcode40. Combination Sum I

---

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

void backTrack(vector<vector<int>>& ans, vector<int>& tmp, const vector<int>& nums, int remain, int step){
        if(remain < 0) return;
        else if(remain == 0) ans.push_back(tmp);
        else{
            for(int i = step; i < nums.size(); i++){
                if(i > step && nums[i - 1] == nums[i]) continue;
                tmp.push_back(nums[i]);
                backTrack(ans, tmp, nums, remain - nums[i], i + 1);
                tmp.pop_back();
            }
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> ans;
        vector<int> tmp;
        sort(candidates.begin(), candidates.end());
        backTrack(ans, tmp, candidates, target, 0);
        return ans;
    }