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]
]
思路
對于nums數組中的每一個數，都依次放入結果集中，如果結果集中已經包含這個數，就繼續下一次循環。
以數組[1,2,3]為例，每次循環的結果是：
[1,2,3]
[1,3,2]
[2,1,3]
[2,3,1]
[3,1,2]
[3,2,1]
代碼
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
backtrack(list, new ArrayList<>(), nums);
return list;
}
private void 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
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]
]
思路
這道題比上一道題多了一個條件，即數組中有重復的數。有兩種思路：
- 仍然按照上一道題的解法，但是把結果用set保存，最終轉換成list。
- 考慮數組中有相同的數，規定
必須按照從前到后的順序使用數字，即數組[1,1]，在組合時，必須先使用第一個1，才能再使用第二個1，這樣就避免了結果集重復的情況。
代碼
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);
}
}
}
78. Subsets
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],
[]
]
思路
與題目Permutations類似，但是Permutations是達到數組長度才將結果保存，而本題目是求子集，任何集合都需要保存。其中for循環的條件也稍微有些變化。
代碼
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack(list, new ArrayList<>(), nums, 0);
return list;
}
private void 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
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],
[]
]
思路
處理重復的數，和上面是一個思路，即只允許用前面的數字。
代碼
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack(list, new ArrayList<>(), nums, 0);
return list;
}
private void 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);
}
}
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]
]
思路
和Subsets是同一個思路，只不過這次不是求子集，而是加上了限制條件：和為指定的值。
代碼
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;
}
private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
if(remain < 0) return;
else if(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 elements
tempList.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]
]
思路
和上一題有兩點不同的地方：
- 數組中可能有重復的數字
- 不能重復利用數組中的數字
解決數組中可能有重復的數字：
if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
解決不能重復利用數組中的數字（上一題中最后是i，而不是i + 1）：
backtrack(list, tempList, nums, remain - nums[i], i + 1);
代碼
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;
}
private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
if(remain < 0) return;
else if(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);
}
}
}
