Combination Sum系列问题

主要使用方法是backtracking。Combination SumGiven a set of candidate numbers (C) 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.Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).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]

解答：为了减少许多不必要的循环过程，应该先把candidates排序，这样当目前遍历到的元素和已经大于target时，就可以不必再访问candidates后面的元素，直接退回到上个选择处进行选择。另外，由于candidates集合中的每个数都可以使用无数次，故每次递归调用都应该从上次加入的元素开始遍历。

public class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> result = new LinkedList<List<Integer>>();
if (candidates == null || candidates.length == 0) {
return result;
}
Arrays.sort(candidates);
List<Integer> temp = new LinkedList<Integer>();
helper(candidates, target, result, temp, 0);
return result;
}
public void helper (int[] candidates, int target, List<List<Integer>> result, List<Integer> temp, int index) {
if (target == 0) {
result.add(new LinkedList<Integer>(temp));
return;
}
for (int i = index; i < candidates.length && candidates[i] <= target; i++) {
temp.add(candidates[i]);
helper(candidates, target - candidates[i], result, temp, i);  //从上次加入的元素开始遍历
temp.remove(temp.size() - 1);
}
}
}

Combination Sum IIGiven 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.Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).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]

解答：与上题一样应该先把candidates排序。这题需要注意的有两点是：1. 每个元素只能用一次，所以每次递归都应该从当前加入元素的下一个元素开始遍历；2. 集合中含有重复元素，所以在每次for循环在candidates挑选元素时，应将已经挑选过的元素过滤（因为加入此元素的结果已经加入结果集合），避免集合中结果出现重复。递归中则不用考虑起始点与上个元素是否相同，因为递归是在前面元素确定的情况下加入下一元素，它们在同一结果中。

public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> rst = new ArrayList<List<Integer>>();
List<Integer> temp = new ArrayList<Integer>();
helper(rst, temp, candidates, target, 0);
return rst;
}
public void helper(List<List<Integer>> rst, List<Integer> temp, int[] candidates, int target, int index) {
if (target == 0) {
rst.add(new ArrayList<Integer>(temp));
return;
}
for (int i = index; i < candidates.length && candidates[i] <= target; i++) {
if (i > index && candidates[i] == candidates[i - 1]) {
continue;
}
temp.add(candidates[i]);
helper(rst, temp, candidates, target - candidates[i], i + 1);  //从下一个元素开始遍历
temp.remove(temp.size() - 1);
}
}
}

Combination Sum IIIFind 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.Ensure that numbers within the set are sorted in ascending order.Example 1:Input: k = 3, n = 7Output:

[[1,2,4]]

Example 2:Input: k = 3, n = 9Output:

[[1,2,6], [1,3,5], [2,3,4]]

解答：可以看做是上题的一种特殊情况， candidates数组中的元素为1到9，且不含重复元素，上题中去重的判断可以去掉。

public class Solution {
public List<List<Integer>> combinationSum3(int k, int n) {
List<List<Integer>> rst = new ArrayList<List<Integer>>();
List<Integer> temp = new ArrayList<Integer>();
helper(rst, temp, k, n, 1);
return rst;
}
public void helper(List<List<Integer>> rst, List<Integer> temp, int k, int n, int number) {
if (k == temp.size() && n == 0) {
rst.add(new ArrayList<Integer>(temp));
return;
}
for (int i = number; i <= 9 && i <= n; i++) {
temp.add(i);
helper(rst, temp, k, n - i, i + 1);
temp.remove(temp.size() - 1);
}
}
}

总结：本题需要注意的是每次递归里面的循环起始点，以及如何避免结果集的重复。