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]
]
题目标签：Array
这道题目和前面那一题本质一摸一样，不同之处是，这题每一个数字只能用一次，之前那题可以无限用；这一题给的array里有重复的，之前那题没有。所以只要之前那题做了，稍微改动一下就可以了。把递归下去的数字index 从i 改成 i + 1，因为这里不需要重复的数字了。再有就是要设一个条件，如果一个重复的数字，不是出现在第一位的话，就跳过它。
Java Solution:
Runtime beats 83.86%
完成日期：07/17/2017
关键词：Array
关键点：Backtracking with sorted array
public class Solution
{
public List<List<Integer>> combinationSum2(int[] candidates, int target)
{
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(candidates);
backtrack(list, new ArrayList<>(), candidates, target, 0); return list;
} public boolean backtrack(List<List<Integer>> list, List<Integer> tempList,
int[] nums, int remain, int start)
{
if(remain < 0) // if remain is 0 or less than 0, meaning the rest numbers are even greater
return false; // therefore, no need to continue the loop, return false
else if(remain == 0)
{
list.add(new ArrayList<>(tempList));
return false;
}
else
{
for(int i=start; i<nums.length; i++)
{
if(i > start && nums[i] == nums[i-1])
continue;
boolean flag;
tempList.add(nums[i]);
flag = backtrack(list, tempList, nums, remain - nums[i], i+1); // i + 1 because we cannot use same number.
tempList.remove(tempList.size() - 1); if(!flag) // if find a sum or fail to find a sum, there is no need to continue
break;// because it is a sorted array with no duplicates, the rest numbers are even greater.
} return true; // return true because previous tempList didn't find a sum or fail a sum
} }
}
参考资料：
http://www.cnblogs.com/grandyang/p/4419386.html
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