Given a set of?distinct?integers,?nums, return all possible subsets.
Note:?The solution set must not contain duplicate subsets.
For example, If?nums?=?[1,2,3], a solution is:
[[3],[1],[2],[1,2,3],[1,3],[2,3],[1,2],[]
] public List<List<Integer>> subsets(int[] nums) {List<List<Integer>> list = new ArrayList<>();Arrays.sort(nums);backtrack(list, new ArrayList<>(), nums, 0);return list;
}privatevoid backtrack(List<List<Integer>> list , List<Integer> tempList, int [] nums, int start){list.add(new ArrayList<>(tempList));for(int i = start; i < nums.length; i++){tempList.add(nums[i]);backtrack(list, tempList, nums, i + 1);tempList.remove(tempList.size() - 1);}
}
?
90 Subsets II (contains duplicates)
Given a collection of integers that might contain duplicates,?nums, return all possible subsets.
Note:?The solution set must not contain duplicate subsets.
For example, If?nums?=?[1,2,2], a solution is:
[[2],[1],[1,2,2],[2,2],[1,2],[]
] public List<List<Integer>> subsetsWithDup(int[] nums) {List<List<Integer>> list = new ArrayList<>();Arrays.sort(nums);backtrack(list, new ArrayList<>(), nums, 0);return list;
}privatevoid backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int start){list.add(new ArrayList<>(tempList));for(int i = start; i < nums.length; i++){if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates tempList.add(nums[i]);backtrack(list, tempList, nums, i + 1);tempList.remove(tempList.size() - 1);}
}
46?Permutations?
Given a collection of?distinct?numbers, return all possible permutations.
For example, [1,2,3]?have the following permutations:
[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]
] public List<List<Integer>> permute(int[] nums) {List<List<Integer>> list = new ArrayList<>();// Arrays.sort(nums); // not necessarybacktrack(list, new ArrayList<>(), nums);return list;
}privatevoid backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums){if(tempList.size() == nums.length){list.add(new ArrayList<>(tempList));} else{for(int i = 0; i < nums.length; i++){ if(tempList.contains(nums[i])) continue; // element already exists, skip tempList.add(nums[i]);backtrack(list, tempList, nums);tempList.remove(tempList.size() - 1);}}
}
47 Permutations II (contains duplicates)?
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example, [1,1,2]?have the following unique permutations:
[[1,1,2],[1,2,1],[2,1,1]
] public List<List<Integer>> permuteUnique(int[] nums) {List<List<Integer>> list = new ArrayList<>();Arrays.sort(nums);backtrack(list, new ArrayList<>(), nums, new boolean[nums.length]);return list;
}private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, boolean [] used){if(tempList.size() == nums.length){list.add(new ArrayList<>(tempList));} else{for(int i = 0; i < nums.length; i++){if(used[i] || i > 0 && nums[i] == nums[i-1] && !used[i - 1]) continue;used[i] = true; tempList.add(nums[i]);backtrack(list, tempList, nums, used);used[i] = false; tempList.remove(tempList.size() - 1);}}
}
39 Combination Sum?
Given a?set?of candidate numbers (C)?(without duplicates)?and a target number (T), find all unique combinations in?C?where the candidate numbers sums to?T.
The?same?repeated number may be chosen from?C?unlimited number of times.
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
?
For example, given candidate set?[2, 3, 6, 7]?and target?7,? A solution set is:?
[[7],[2, 2, 3]
] public List<List<Integer>> combinationSum(int[] nums, int target) {List<List<Integer>> list = new ArrayList<>();Arrays.sort(nums);backtrack(list, new ArrayList<>(), nums, target, 0);return list;
}privatevoid backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){if(remain < 0) return;elseif(remain == 0) list.add(new ArrayList<>(tempList));else{ for(int i = start; i < nums.length; i++){tempList.add(nums[i]);backtrack(list, tempList, nums, remain - nums[i], i); // not i + 1 because we can reuse same elementstempList.remove(tempList.size() - 1);}}
}
40 Combination Sum II (can't reuse same element)?
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in?C?where the candidate numbers sums to?T.
Each number in?C?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.
?
For example, given candidate set?[10, 1, 2, 7, 6, 1, 5]?and target?8,? A solution set is:?
[[1, 7],[1, 2, 5],[2, 6],[1, 1, 6]
] public List<List<Integer>> combinationSum2(int[] nums, int target) {List<List<Integer>> list = new ArrayList<>();Arrays.sort(nums);backtrack(list, new ArrayList<>(), nums, target, 0);return list;}privatevoid backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){if(remain < 0) return;elseif(remain == 0) list.add(new ArrayList<>(tempList));else{for(int i = start; i < nums.length; i++){if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates tempList.add(nums[i]);backtrack(list, tempList, nums, remain - nums[i], i + 1);tempList.remove(tempList.size() - 1); }}
}
131 Palindrome Partitioning
Given a string?s, partition?s?such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of?s.
For example, given?s?=?"aab", Return
[["aa","b"],["a","a","b"]
] public List<List<String>> partition(String s) {List<List<String>> list = new ArrayList<>();backtrack(list, new ArrayList<>(), s, 0);return list;
}publicvoid backtrack(List<List<String>> list, List<String> tempList, String s, int start){if(start == s.length())list.add(new ArrayList<>(tempList));else{for(int i = start; i < s.length(); i++){if(isPalindrome(s, start, i)){tempList.add(s.substring(start, i + 1));backtrack(list, tempList, s, i + 1);tempList.remove(tempList.size() - 1);}}}
}publicboolean isPalindrome(String s, int low, int high){while(low < high)if(s.charAt(low++) != s.charAt(high--)) returnfalse;returntrue;
}
