/*排列与组合专题：
排列与元素的顺序有关
组合与顺序无关
permutation: 排列
combination: 组合 在组合之前，需要排序
*/
public class cp {
/*
LeetCode 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) {
// variable of result
List<List<Integer>> result = new ArrayList<>();
// corner
if(nums.length == 0 || nums == null) {
return result;
}
// sub-result
List<Integer> list = new ArrayList<>();
// recursive
helper(nums, result, list);
// return
return result;
}
private void helper (int[] nums, List<List<Integer>> result, List<Integer> list) {
// exit
if(list.size == nums.length) {
result.add(new ArrayList(list));
return;
}
// for loop
for(int i = 0; i < nums.length; i++) {
if(list.contains(nums[i])) {
continue;
}
// add
list.add(nums[i]);
// recursive
helper(nums, result, list);
// resert
list.remove(list.size()-1);
}
}
/*
LeetCode 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]
]
*/
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
if(nums.length == 0 || nums == null) {
return result;
}
List<Integer> list = new ArrayList<>();
int[] visited = new int[nums.length];
Arrays.sort(nums);
helper(nums, result, list, visited);
return result;
}
private void helper(int[] nums, List<List<Integer>> result,
List<Integer> list, int[] visited) {
if(list.size() == nums.length) {
result.add(new ArrayList(list));
return;
}
for(int i = 0; i < nums.length; i ++) {
if(visited[i] == 1 || (i != 0 && nums[i] == nums[i-1] && visited[i-1] == 0)) {
continue;
}
visited[i] = 1;
list.add(nums[i]);
helper(nums, result, list, visited);
list.remove(list.size() - 1);
visited[i] = 0;
}
}
/*
LeetCode 77. Combinations
Given two integers n and k,
return all possible combinations of k numbers out of 1 ... n.
For example, if n = 4 and k = 2, a solution is:
[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]
*/
public List<List<Integer>> combinations (int n, int k) {
List<List<Integer>> result = new ArrayList<>();
if(n < k || n == 0 || k == 0) {
return result;
}
List<Integer> solution = new ArrayList<>();
helper(n, k, 1, result, solution);
return result;
}
private void helper(int n, int k, int start,
List<List<Integer>> result,
List<Integer> solution) {
if(solution.size() == k) {
result.add(new ArrayList(solution));
return;
}
for(int i = start; i <= n; i ++) {
solution.add(i);
helper(n, k, i + 1, result, solution);
solution.remove(solution.size() - 1);
}
}
/*
LeetCode 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[] candidates, int target) {
List<List<Integer>> result = new ArrayList<>();
if (candidates == null || candidates.length == 0) {
return result;
}
List<Integer> combination = new ArrayList<>();
Arrays.sort(candidates);
dfs(candidates, target, 0, combination, result);
return result;
}
public static void dfs(int[] candidates, int target, int index,
List<Integer> combination,
List<List<Integer>> result) {
if (target == 0) {
result.add(new ArrayList<Integer>(combination));
return;
}
for (int i = index; i < candidates.length; i ++) {
if (candidates[i] > target) {
break;
}
// if (i != index && candidates[i] == candidates[i-1]) {
// continue;
// }
combination.add(candidates[i]);
dfs(candidates, target - candidates[i], i, combination, result);
combination.remove(combination.size()-1);
}
}
/*
40. Combination Sum II
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[] candidates, int target) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> combination = new ArrayList<>();
if (candidates == null) {
return result;
}
Arrays.sort(candidates);
dfs(candidates, target, 0, combination, result);
return result;
}
public static void dfs(int[] candidates, int target, int index,
List<Integer> combination,
List<List<Integer>> result) {
if (target == 0) {
result.add(new ArrayList<Integer>(combination));
return;
}
for (int i = index; i < candidates.length; i ++) {
if (target < candidates[i]) {
break;
}
if (i != index && candidates[i] == candidates[i-1]) {
continue;
}
combination.add(candidates[i]);
dfs(candidates, target - candidates[i], i + 1, combination, result);
combination.remove(combination.size()-1);
}
}
/*
216. Combination Sum III
Find all possible combinations of k numbers
that add up to a number n,
given that only numbers from 1 to 9 can be used
and each combination should be a unique set of numbers.
Example 1: Input: k = 3, n = 7
Output: [[1,2,4]]
Example 2: Input: k = 3, n = 9
Output: [[1,2,6], [1,3,5], [2,3,4]]
*/
public List<List<Integer>> combinationSum3(int k, int n) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> combination = new ArrayList<>();
if (k > n || n == 0 || k == 0) {
return result;
}
dfs(k, n, 1, result, combination);
return result;
}
public void dfs(int k, int n, int index, List<List<Integer>> result, List<Integer> combination) {
if (n == 0 && combination.size() == k) {
result.add(new ArrayList<Integer>(combination));
return;
}
for (int i = index; i <= 9; i ++) {
combination.add(i);
dfs(k, n-i, i+1, result, combination);
combination.remove(combination.size()-1);
}
}
}
网友评论