给定一个候选人编号的集合 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的每个数字在每个组合中只能使用 一次 。
注意：解集不能包含重复的组合。
示例 1:
输入: candidates = [10,1,2,7,6,1,5], target = 8, 输出: [ [1,1,6], [1,2,5], [1,7], [2,6] ]
示例 2:
输入: candidates = [2,5,2,1,2], target = 5, 输出: [ [1,2,2], [5] ]
提示:
1 <= candidates.length <= 1001 <= candidates[i] <= 501 <= target <= 30
思路：和39一样的思路，注意回溯
class Solution {List<List<Integer>> result = new ArrayList<>();LinkedList<Integer> path = new LinkedList<>();public List<List<Integer>> combinationSum2(int[] candidates, int target) {if (candidates == null || candidates.length == 0){return result;}Arrays.sort(candidates);backTrachking(candidates,target,0);return result;}public void backTrachking(int[] candidates,int target,int index){//终止条件if (target == 0){result.add(new ArrayList<>(path));return;}if (target < 0){return;}if (index == candidates.length){return;}for (int i = index; i < candidates.length; i++) {//正确剔除重复解的办法//跳过同一树层使用过的元素if ( i > index && candidates[i] == candidates[i - 1] ) {continue;}int curNum = candidates[i];if (target - curNum < 0){break;}path.add(curNum);backTrachking(candidates,target - curNum,i + 1);//这里加一是不能重复使用path.removeLast();}}
}
