leetcode中常見(jiàn)的backTrack類(lèi)題目：combination、subsets、permutation、Palindrome Partitioning.

1、combination

<1>leetcode39. Combination Sum

題目描述：

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

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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;}

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